NCERT Solutions Class 11 Maths Chapter 11

NCERT Solutions Class 11 Mathematics Chapter 11 – Conic sections

Mathematics has a wide range of uses in many spheres of education. Algebra and Geometry are amongst the most important part of Mathematics. It is the study of equations and their applications, statistics and their representation, graphs and their plotting, theorems and their assumptions and numbers and their calculations. All these are highly required to master this subject. Hence, the subject holds a lot of importance.

Conic sections are one of the crucial chapters of Class 11 Mathematics as it carries a lot of weightage in the examinations. The topics covered in this chapter include circle, Ellipse, parabola, hyperbola, degenerate conic sections, standard equations of such curves, the relationship between semi-major axis, semi-minor axis, and the distance of the focus from the centre of the Ellipse, as well as exceptional cases of the Ellipse, eccentricity, and latus rectum.

The Class 11 Mathematics Chapter 11 is difficult so it is required to have access to suitable study material. As a result, Extramarks ensures that the study material provided by it proves beneficial for its students. In our NCERT Solutions Class 11 Mathematics Chapter 11, we have covered all the answers to challenging questions in a simplified way making this chapter easy to understand and interesting for the students.

If you are searching for quality resources, Extramarks is the right platform to rely upon. Not only is our study material well known, but also the mock tests provided by us have always had a remarkable impact on the students’ results. One can get NCERT solutions Class 11 Mathematics Chapter 11 from the Extramarks website by registering on it.

Key Topics Covered In Conic sections Class 11 Mathematics Chapter 11

The Conic section is the study of the various segments of geometry like circles, Ellipses, parabolas and Hyperbola. Students learn how to use different parameters of these geometric shapes in doing calculations using graphs. It eases the methods used in long calculations helping students to achieveaccuracy. The various topics and concepts covered in this chapter will be used in the higher studies of Mathematics. As a result, students must grasp all the concepts covered in the conic section entirely and with proper understanding.

Every core topic of the chapter is highlighted in the NCERT Solutions Class 11 Mathematics Chapter 11 in a manner that students get a clear understanding of it. Along with it, students can also find solved examples and answers to the textbook exercise Through all this, they can  practice various questions on each concept to gain a better understanding of the chapter. This helps them to test themselves right at the moment they have completed the chapter. They can get the NCERT Solutions from the Extramarks’ website.

NCERT Solutions Class 11 Mathematics Chapter 11 requires students to think more broadly and enhance their understanding of Mathematics in an accurate manner

Introductions

In the earlier chapter, we studied the equation of a line. In this chapter, we will study Curves, Circles, Ellipses, Parabola, and Hyperbola.

These Curves, Circles, Ellipse, Parabola, and Hyperbola are known as Conic sections or conics because they can be obtained as intersections of a plane with a double-napped right circular cone.

Sections of a cone

In the section of this chapter, we will be talking about the sections of the cone. We assume L as a fixed vertical line, and another line intersects at a fixed point v, which is inclined to attend the angle of α and β. It is the angle formed by the intersection of the plane and the vertical axis of the cone.

• Vertex

It is the point of intersection of the conic section and axis.

• Nappes

Nappes are the points where the vertex separates the cone into two parts.

• Conic sections

The curves of a Circle, Ellipse, Parabola and Hyperbola are known as Conic sections or conics.

Circle, Ellipse, Parabola and Hyperbola

When the plane intersects the nappe other than the vertex of the cone, we have the following situations:

(a) When β = 90°, the section is a circle

(b) When α < β < 90°, the section is an Ellipse

(c) When β = α; the section is a parabola

(In each of the above three situations, the plane intersects entirely across one nappe of the cone).

(d) When 0 ≤ β < α, the plane intersects through both the nappes and the curves of intersection is a hyperbola

Degenerated conic section.

When the plane intersects at the vertex of the cone, we have the following different cases:

(a) If α < β ≤ 90°, then the section is a point

(b) If β = α, the plane contains a generator of the cone, and the section is a straight line

If the fixed point lies on the fixed-line, then the set of points in the plane, which are equidistant from the fixed point and the fixed-line, is the straight line through the fixed point and perpendicular to the fixed-line. We call this straight line a degenerate case of the parabola.

This is the degenerated case of a parabola.

(c). If 0 < β < α, the section is a pair of intersecting straight lines. It is the degenerated case of a hyperbola.

Circle

A circle is a set of all the definite points which are equidistant from a fixed point in two dimensional plane.

• The fixed point is known as the centre of the circle
• distance from the centre to a point on the circle is known as the radius of the circle

The equation of a circle from the centre (h,k) and the radius r is given by:

[ (x-h)2+(y-k)2] = r

i.e. (x – h) 2 + (y – k) 2 = r 2

For more details about the circle, students can visit the website of Extramarks and refer to NCERT Solutions Class 11 Mathematics Chapter 11

Parabola

A parabola is the collection of all the definite points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane.

• The fixed-line is called the directrix of the parabola
• The fixed point F is known as the focus.
• A line passing through the focus and perpendicular to the directrix is called the axis of the parabola.
• The point which intersects with the parabola and the axis is called the vertex of the parabola.

Standard equation of the parabola

The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along either of the x-axis or y-axis.

The standard equation to the parabola with vertex at the origin, focus at (a,0), and directrix x = -a is given by

y2 = 4ax

From the standard equations of the parabola, we draw the following observations:

• Parabola is symmetric with respect to the axis of the parabola. If the equation has a y2 term, then the axis of symmetry is along the x-axis, and if the equation has a x2 term, then the axis of symmetry is along the y-axis.
• When the axis of symmetry is along the x-axis, the parabola opens to the
• right side if the coefficient of x is +ve
• left side if the coefficient of x is -ve.
• When the axis of symmetry is along the y-axis, the parabola opens
• Upwards if the coefficient of the y is +ve.
• Downwards if the coefficient of the y is -ve.

• Latus rectum

The Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose endpoints lie on the parabola.

The length of the latus rectum is 4a.

Ellipse

An Ellipse is the set of all the definite points in a plane, the sum of whose distances from two fixed points in the plane is a constant.

• The two points that are fixed are known as the foci of the Ellipse.
• The mid point of the line segment that joined to the foci is known as the Ellipse’s centre.
• The line segment passing from the foci of the Ellipse is known as the major axis
• The line segment passing from the centre and perpendicular to the major axis is known as the minor axis.
• The endpoints containing the major axis are known as the vertices of the Ellipse.

Relationship between the semi-minor axis, semi-major axis and the distance of the focus from the centre of the Ellipse

By definition of an Ellipse, we give the relationship between the semi-major axis, semi-minor axis and the distance of the focus from the centre of the Ellipse. It is given as

a = √(b2+c2).

Special cases of the Ellipse

In the equation above, if we keep fixed and vary c from 0 to a, the resulting Ellipses will vary in   shape.

Case 1: When c = 0, both foci merge with the centre of the Ellipse and a2 = b2, i.e., a = b, and so the Ellipse becomes a circle. Hence, the circle is a special case of an Ellipse

Case 2: If c = a, then b = 0. The Ellipse derives the line segment F1F2, joining the two foci.

Eccentricity,

The eccentricity of an Ellipse, is the proportion of the distances from the centre of the Ellipse to one of the foci and one of the vertices of the Ellipse.

It is denoted by ‘e’ i.e,

e= c/a

Standard equation of an Ellipse.

The equation of an ellipse is easiest if the ellipse’s centre is at the origin and the foci are on the x-axis or y-axis.

The equation of the Ellipse is given as

x2/b2 + y2/a2= 1

Or

x2/a + y2/b= 1.

These two equations are known as standard equations of the ellipses.

We have the following observations for the standard equations of the ellipses:

• Ellipse is symmetric with relation to both the coordinate axis since if (x, y) is a point on the Ellipse, then (– x, y), (x, –y) and (– x, –y) are also all points on the ellipse.
• The foci are always present on the major axis. The major axis can be marked by finding the intercepts on the axis of symmetry, i.e. the major axis is along the y-axis if the coefficient of y2 has the larger denominator, and it is along the x-axis if the coefficient of x2 has the larger denominator.

Latus rectum

The Ellipse’s latus rectum is a line segment perpendicular to the major axis through any foci whose endpoints lie on the Ellipse.

The length of the latus rectum in a hyperbola is 2b2 / a

Hyperbola

The hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

• The term “difference” used in the definition means the distance to the farther point minus the distance to the closest point.
• The two fixed points which are known as the foci of the hyperbola.
• The mid-point of the line segment connecting the foci is called the hyperbola’s centre.
• The line passing through the foci is called the transverse axis.
• The line passing from the centre and perpendicular to the transverse axis is known as the conjugate axis.
• The points where the hyperbola intersects the transverse axis are called the hyperbola’s vertices.

Eccentricity

As an Ellipse, the ratio e = c / a is known as the eccentricity of the hyperbola. If c ≥ a the eccentricity is more than one.

In eccentricity, the foci are presant at a distance of ‘a.e’ from the centre.

Standard equation of hyperbola

The equation of a hyperbola is easiest if the centre of the hyperbola is at the origin and the foci are on the x-axis or y-axis.

The equation of the hyperbola is given by:

y2/a2 – x2/b2 = 1.

This equation is known as the standard equation of hyperbola

A hyperbola in which a = b is known as an equilateral hyperbola.

From the standard equations of hyperbolas, we have the following observations:

• Hyperbola is symmetric for both the axes. if the point (x, y) is on the hyperbola, the point (– x, y), (x, – y) and (– x, – y) are also on the hyperbola.
• The foci are always on the transverse axis. It is having the positive term whose denominator gives the transverse axis.

Latus rectum

The Latus rectum of the hyperbola is a line segment perpendicular to the transverse axis through any foci whose end points lie on the hyperbola. The length of the latus rectum in hyperbola is 2.b2/a

Students can access our NCERT Solutions Class 11 Mathematics Chapter 11 by registering on Extramarks’ website. Our solutions include chapter-specific study notes, important questions and solutions, revision notes, etc. all which has proven beneficial to lakhs of students.

NCERT Solutions Class 11 Mathematics Chapter 11 Exercise & Solutions

NCERT solutions Class 11 Mathematics Chapter 11 has a detailed answers to the texbook exercises, solved examples and additional questions for practice. You can level up your preparation and improve your performance in the examinations by referring to it.

The subject matter experts have given the solutions. All the solutions are in a step-by-step format, including every single point required while calculating every answer. Experienced teachers from the field of Mathematics have curated our solutions given in the study materials. Hence, you can wholeheartedly trust our study resources.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions Class 11 Mathematics Chapter 11:

• Chapter 11: Exercise 11.1
• Chapter 11: Exercise 11.2
• Chapter 11: Exercise 11.3
• Chapter 11: Exercise 11.4

Along with Class 11 Mathematics solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

NCERT Exemplar Class 11 Mathematics

NCERT Exemplar Class 11 Mathematics book covers complex theories that guides the students to prepare and face the upcoming examinations confidently. This book helps students develop confidence during their preparation as they have questions of basic as well as advanced level. Students can refer to it for additional practice.

The questions are designed as per the pattern of the CBSE as well as competitive examinations. This will aid students in solving all types of questions with accuracy, helping them develop an interest in the subject at large. Our solutions have always proven to be a remarkable guide for the students preparing for their school examinations and other competitive examinations like NEET, JEE, WBJEE, BITSAT, MHT-CET etc.

Students can ensure guaranteed success once they include NCERT solutions Class 11 Mathematics Chapter 11 and NCERT Exemplar Class 11 Mathematics in their study material. Both of these resources can be easily availed from the Extramarks’ website.

Key Features for NCERT Solutions Class 11 Mathematics Chapter 11

To score well, you must revise the concepts repeatedly. Hence, NCERT Solutions for Class 11 Mathematics Chapter 11 offers guides to revise appropriately. The key features are as follows:

• The revision-related material can be accessed from our NCERT Solutions Class 11 Mathematics Chapter 11.
• It has concepts divided into various segments, helping students get an in-depth understanding of the chapters .
• After completing the NCERT Solutions for Class 11 Mathematics Chapter 11, students can easily understand the interrelations between the various geometric shapes in their higher education.

Q.1 In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
2. centre (–2,3) and radius 4
3.

$\mathbf{centre}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\right)}\mathbf{\text{}}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\frac{\mathbf{1}}{\mathbf{12}}$

4.

$\mathbf{centre}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{\right)}\mathbf{\text{}}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{\mathbf{2}}$

5.

$\mathbf{centre}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{-}\mathbf{a}\mathbf{,}\mathbf{-}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}$

Ans

1.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(0,2\right)\text{and radius 2 is:}\\ {\left(\mathrm{x}-0\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}={2}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{y}+4=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

2.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(-2,3\right)\text{and radius 4 is:}\\ {\left(\mathrm{x}+2\right)}^{2}+{\left(\mathrm{y}-3\right)}^{2}={4}^{2}\\ ⇒{\mathrm{x}}^{2}+4\mathrm{x}+4+{\mathrm{y}}^{2}-6\mathrm{y}+9=16\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+4\mathrm{x}-6\mathrm{y}-3=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

3.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(\frac{1}{2},\frac{1}{4}\right)\text{and radius}\frac{1}{12}\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\frac{1}{2}\right)}^{2}+{\left(\mathrm{y}-\frac{1}{4}\right)}^{2}={\left(\frac{1}{12}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}+\frac{1}{4}+{\mathrm{y}}^{2}-\frac{1}{2}\mathrm{y}+\frac{1}{16}=\frac{1}{144}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}+{\mathrm{y}}^{2}-\frac{1}{2}\mathrm{y}+\frac{1}{4}+\frac{1}{16}-\frac{1}{144}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}+{\mathrm{y}}^{2}-\frac{1}{2}\mathrm{y}+\frac{11}{36}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}36{\mathrm{x}}^{2}+36{\mathrm{y}}^{2}-36\mathrm{x}-18\mathrm{y}+11=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

4.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(1,1\right)\text{and radius}\sqrt{\text{2}}\text{is:}\end{array}$ $\begin{array}{}\end{array}$ $\begin{array}{l}{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-1\right)}^{2}={\left(\sqrt{\text{2}}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-2\mathrm{x}+1+{\mathrm{y}}^{2}-2\mathrm{y}+1=2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-2\mathrm{x}-2\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

5.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(-\mathrm{a},-\mathrm{b}\right)\text{and radius}\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^{2}}\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}+\mathrm{a}\right)}^{2}+{\left(\mathrm{y}+\mathrm{b}\right)}^{2}={\left(\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^{2}}\right)}^{2}\\ {\mathrm{x}}^{2}+2\mathrm{ax}+{\mathrm{a}}^{2}+{\mathrm{y}}^{2}+2\mathrm{by}+{\mathrm{b}}^{2}={\text{a}}^{\text{2}}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+2\mathrm{ax}+2\mathrm{by}+2{\mathrm{b}}^{2}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.2 In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)2 + (y – 3)2 = 36
7. x2 + y2 – 4x – 8y – 45 = 0
8. x2 + y2 – 8x + 10y – 12 = 0
9. 2x2 + 2y2 – x = 0

Ans

6.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{(x+5)}}^{\text{2}}\text{+(y}-{\text{3)}}^{\text{2}}\text{=36}...\left(\text{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=-5,\mathrm{k}=3{\text{and r}}^{\text{2}}=\text{36}\end{array}$ $\begin{array}{l}⇒\mathrm{h}=-5,\mathrm{k}=3\text{and r}=\text{6}\\ \text{Therefore, centre is}\left(-5,3\right)\text{and radius is 6.}\end{array}$

7.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{x}-8\mathrm{y}-45=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-4\mathrm{x}\right)+\left({\mathrm{y}}^{2}-8\mathrm{y}\right)-45=0\\ ⇒\left({\mathrm{x}}^{2}-4\mathrm{x}+{2}^{2}-{2}^{2}\right)+\left({\mathrm{y}}^{2}-8\mathrm{y}+{4}^{2}-{4}^{2}\right)-45=0\\ ⇒\left({\mathrm{x}}^{2}-4\mathrm{x}+{2}^{2}\right)-4+\left({\mathrm{y}}^{2}-8\mathrm{y}+{4}^{2}\right)-16-45=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-2\right)}^{2}+{\left({\mathrm{y}}^{2}-4\right)}^{2}-65=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-2\right)}^{2}+{\left({\mathrm{y}}^{2}-4\right)}^{2}={\left(\sqrt{65}\right)}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=2,\mathrm{k}=4\text{and r}=\sqrt{65}\\ \text{Therefore, centre is}\left(2,4\right)\text{and radius is}\sqrt{\text{65}}\text{.}\end{array}$

8.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-8\mathrm{x}+10\mathrm{y}-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-8\mathrm{x}\right)+\left({\mathrm{y}}^{2}+10\mathrm{y}\right)-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-8\mathrm{x}+{4}^{2}-{4}^{2}\right)+\left({\mathrm{y}}^{2}+10\mathrm{y}+{5}^{2}-{5}^{2}\right)-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-8\mathrm{x}+{4}^{2}\right)-16+\left({\mathrm{y}}^{2}+10\mathrm{y}+{5}^{2}\right)-25-12=0\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-4\right)}^{2}+{\left({\mathrm{y}}^{2}+5\right)}^{2}-53=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-4\right)}^{2}+{\left({\mathrm{y}}^{2}+5\right)}^{2}={\left(\sqrt{53}\right)}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=4,\mathrm{k}=-5\text{and r}=\sqrt{53}\\ \text{Therefore, centre is}\left(4,-5\right)\text{and radius is}\sqrt{\text{53}}\text{.}\end{array}$

9.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}2x}}^{\text{2}}+{\text{2y}}^{\text{2}}-\text{x}=0\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}-\frac{1}{\text{2}}\text{x}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-\frac{1}{2}\mathrm{x}\right)+{\mathrm{y}}^{2}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-\frac{1}{2}\mathrm{x}+\frac{1}{{4}^{2}}-\frac{1}{{4}^{2}}\right)+{\mathrm{y}}^{2}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-\frac{1}{2}\mathrm{x}+\frac{1}{{4}^{2}}\right)+{\mathrm{y}}^{2}=\frac{1}{{2}^{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\frac{1}{4}\right)}^{2}+{\mathrm{y}}^{2}=\frac{1}{{4}^{2}}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{1}{4},\mathrm{k}=0\text{and r}=\frac{1}{4}\end{array}$ $\text{Therefore, centre is}\left(\frac{1}{4},0\right)\text{and radius is}\frac{1}{4}\text{.}$

Q.3 Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{\hspace{0.17em}}\left(\mathrm{i}\right)\text{is passing through}\left(4,1\right)\text{and}\left(6,5\right)\text{then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(4-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{ii}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}{\left(6-\mathrm{h}\right)}^{2}+{\left(5-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}and equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(4-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}={\left(6-\mathrm{h}\right)}^{2}+{\left(5-\mathrm{k}\right)}^{2}\\ ⇒16-8\mathrm{h}+{\mathrm{h}}^{2}+1-2\mathrm{k}+{\mathrm{k}}^{2}=36-12\mathrm{h}+{\mathrm{h}}^{2}+25-10\mathrm{k}+{\mathrm{k}}^{2}\\ ⇒4\mathrm{h}+8\mathrm{k}=44\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}+2\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{centre}\left(\mathrm{h},\mathrm{k}\right)\text{lies on the line 4x}+\mathrm{y}=16.\\ \mathrm{So},4\mathrm{h}+\mathrm{k}=16...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\text{3 and k}=\text{4}\\ \text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(4-3\right)}^{2}+{\left(1-4\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{2}=1+9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\end{array}$ $\begin{array}{l}\mathrm{Then},\text{equation of circle is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-4\right)}^{2}=10\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-8\mathrm{y}+16=10\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-6\mathrm{x}-8\mathrm{y}+15=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.4 Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{\hspace{0.17em}}\left(\mathrm{i}\right)\text{is passing through}\left(2,3\right)\text{and}\left(-1,1\right)\text{then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2-\mathrm{h}\right)}^{2}+{\left(3-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{ii}\right)\\ \mathrm{and}\text{\hspace{0.17em}}{\left(-1-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}and equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2-\mathrm{h}\right)}^{2}+{\left(3-\mathrm{k}\right)}^{2}={\left(-1-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4-4\text{\hspace{0.17em}}\mathrm{h}+{\mathrm{h}}^{2}+9-6\mathrm{k}+{\mathrm{k}}^{2}=1+2\mathrm{h}+{\mathrm{h}}^{2}+1-2\mathrm{k}+{\mathrm{k}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}}6\mathrm{h}-4\mathrm{k}=-11\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}6\text{\hspace{0.17em}}\mathrm{h}+4\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{centre}\left(\mathrm{h},\mathrm{k}\right)\text{lies on the line x}-3\mathrm{y}=11.\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}-3\mathrm{k}=11...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\frac{7}{2}\text{and k}=-\frac{\text{5}}{2}\end{array}$ $\begin{array}{l}\text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {\left(2-\frac{7}{2}\right)}^{2}+{\left(3+\frac{\text{5}}{2}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{2}=\frac{9}{4}+\frac{121}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{130}{4}\\ \mathrm{Then},\text{equation of circle is:}\\ {\left(\mathrm{x}-\frac{7}{2}\right)}^{2}+{\left(\mathrm{y}-\frac{\text{5}}{2}\right)}^{2}=\frac{130}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-7\mathrm{x}+\frac{49}{4}+{\mathrm{y}}^{2}+5\mathrm{y}+\frac{25}{4}=\frac{130}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{x}}^{2}+4{\mathrm{y}}^{2}-28\mathrm{x}+20\mathrm{y}=56\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-7\mathrm{x}+5\mathrm{y}-14=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.5 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{a},0\right)\text{and radius 5 is:}\\ {\left(\mathrm{x}-\mathrm{a}\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\mathrm{a}\right)}^{2}+{\mathrm{y}}^{2}={5}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{circle}\left(\mathrm{i}\right)\text{passes through the point}\left(2,3\right).\text{\hspace{0.17em}}\mathrm{So},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2-\mathrm{a}\right)}^{2}+{3}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4-4\mathrm{a}+{\mathrm{a}}^{2}+9=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-4\mathrm{a}=25-13\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-4\mathrm{a}-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{a}-6\right)\left(\mathrm{a}+2\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=6,-2\\ \mathrm{Case}\text{\hspace{0.17em}}1:\mathrm{When}\text{centre of circle is}\left(6,0\right):\\ \mathrm{The}\text{equation of circle is:-}\\ {\left(\mathrm{x}-6\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-12\mathrm{x}+36+{\mathrm{y}}^{2}=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-12\mathrm{x}+{\mathrm{y}}^{2}+11=0\\ \mathrm{Case}\text{\hspace{0.17em}}2:\mathrm{When}\text{centre of circle is}\left(-2,0\right):\\ \mathrm{The}\text{equation of circle is:-}\\ {\left(\mathrm{x}+2\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+4\mathrm{x}+4+{\mathrm{y}}^{2}=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+4\mathrm{x}+{\mathrm{y}}^{2}-21=0\\ \mathrm{Thus},\text{the equation of circle is}\\ {\mathrm{x}}^{2}+{\mathrm{y}}^{2}+4\mathrm{x}-21=0\text{or}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-12\mathrm{x}+11=0.\end{array}$

Q.6 Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the points}\left(0,0\right),\left(\mathrm{a},0\right)\text{and}\left(0,\mathrm{b}\right),\text{so}\\ {\left(0-\mathrm{h}\right)}^{2}+{\left(0-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{r}}^{2}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}{\left(\mathrm{a}-\mathrm{h}\right)}^{2}+{\left(0-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-2\mathrm{ah}+{\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iii}\right)\\ {\left(0-\mathrm{h}\right)}^{2}+{\left(\mathrm{b}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}+{\mathrm{b}}^{2}-2\mathrm{bk}+{\mathrm{k}}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iv}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we get}\\ {\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{a}}^{2}-2\mathrm{ah}+{\mathrm{h}}^{2}+{\mathrm{k}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-2\mathrm{ah}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}\left(\mathrm{a}-2\mathrm{h}\right)=0\\ ⇒\text{\hspace{0.17em}}\mathrm{a}-2\mathrm{h}=0,\mathrm{a}\ne 0\\ ⇒\text{\hspace{0.17em}}\mathrm{h}=\frac{\mathrm{a}}{2}\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iv}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{h}}^{2}+{\mathrm{b}}^{2}-2\mathrm{bk}+{\mathrm{k}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}-2\mathrm{bk}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}\left(\mathrm{b}-2\mathrm{k}\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}-2\mathrm{k}=0,\mathrm{b}\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{\mathrm{b}}{2}\\ \mathrm{Putting}\text{values of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\frac{\mathrm{b}}{2}\right)}^{2}={\mathrm{r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{a}}^{2}}{4}+\frac{{\mathrm{b}}^{2}}{4}={\mathrm{r}}^{2}\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right):\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}{\left(\mathrm{x}-\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\mathrm{y}-\frac{\mathrm{b}}{2}\right)}^{2}=\frac{{\mathrm{a}}^{2}}{4}+\frac{{\mathrm{b}}^{2}}{4}\\ ⇒{\mathrm{x}}^{2}-\mathrm{ax}+\frac{{\mathrm{a}}^{2}}{4}+{\mathrm{y}}^{2}-\mathrm{by}+\frac{{\mathrm{b}}^{2}}{4}=\frac{{\mathrm{a}}^{2}}{4}+\frac{{\mathrm{b}}^{2}}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-\mathrm{ax}-\mathrm{by}=0\\ \mathrm{This}\text{is the required equation of the circle.}\end{array}$

Q.7 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(2,2\right)\text{and radius r is:}\\ {\left(\mathrm{x}-2\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the point}\left(4,5\right),\text{so}\\ {\left(4-2\right)}^{2}+{\left(5-2\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4+9={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}13={\mathrm{r}}^{2}\\ \mathrm{Putting}{\text{value of r}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ {\left(\mathrm{x}-2\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}=13\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-4\mathrm{x}+4+{\mathrm{y}}^{2}-4\mathrm{y}+4=13\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{x}-4\mathrm{y}=5\\ \mathrm{This}\text{is the equation of required circle.}\end{array}$

Q.8 Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Ans

$\begin{array}{l}\mathrm{The}\text{equation of given circle is:}\\ {\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{25}\end{array}$ $\begin{array}{l}\text{Putting x}=\text{2}.\text{5},\text{y}=\text{3}.\text{5 in L.H.S. of given equation, we get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{\hspace{0.17em}}{\left(\text{2}.\text{5}\right)}^{\text{2}}+{\left(\text{3}.\text{5}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6.25+12.25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=18.50<25=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{point}\left(–\text{2}.\text{5},\text{3}.\text{5}\right)\text{lies inside of the circle.}\\ \text{As distance of the point from centre is less than radius}\\ \text{of the circle.}\end{array}$

Q.9 In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

1. y2 = 12x
2. x2 = 6y
3. y2 = – 8x
4. x2 = – 16y
5. y2 = 10x
6. x2 = – 9y

Ans

1.

The given equation of parabola is:
y2 = 12x
Since, coefficient of x is positive. So, comparing it with y2 = 4ax and we get
4a = 12
a = 3
The coordinates of the focus is (3, 0).
Since, this equation has y2, so axis is x-axis.
Equation of directrix is: x = – 3.
Length of latus rectum = 4a
= 4 x 3
= 12

2.

The given equation of parabola is: x2 = 6y
Since, coefficient of y is positive. So, comparing it with x2 = 4ay and we get
4a = 6
a = (3/2)
The coordinates of the focus is (0, 3/2).
Since, this equation has x2, so axis is y-axis.
Equation of directrix is: y = – 3/2.
Length of latus rectum = 4a
= 4 x 3/2
= 6

3.

The given equation of parabola is:
y2 = – 8x
Since, coefficient of x is negative. So, comparing it with y2 = – 4ax and we get
– 4a = – 8
a = 2
The coordinates of the focus is (– 2, 0).
Since, this equation has y2, so axis is x-axis.
Equation of directrix is:
x = 2.
Length of latus rectum = 4a
= 4 x 2
= 8

4.

The given equation of parabola is:
x2 = – 16y
Since, coefficient of y is negative. So, comparing it with x2 = – 4ay and we get
4a = – 16
a = –4
The coordinates of the focus is (0, –4).
Since, this equation has x2, so axis is y-axis.
Equation of directrix is: y = 4.
Length of latus rectum = 4a
= 4 x 4
= 16

5.

The given equation of parabola is:
y2 = 10x
Since, coefficient of x is positive. So, comparing it with y2 = 4ax and we get
4a = 10
a = (5/2)
The coordinates of the focus is (5/2, 0).
Since, this equation has y2, so axis is x-axis.
Equation of directrix is: x = – 5/2.
Length of latus rectum = 4a
= 4 x 5/2
= 10

6.

The given equation of parabola is:
x2 = – 9y
Since, coefficient of y is negative. So, comparing it with x2 = – 4ay and we get
4a = – 9
a = – 9/4
The coordinates of the focus is (0, – 9/4).
Since, this equation has x2, so axis is y-axis.
Equation of directrix is: y = 9/4.
Length of latus rectum = 4a
= 4 x 9/4
= 9

Q.10 In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
7. Focus (6,0); directrix x = – 6
8. Focus (0,–3); directrix y = 3
9. Vertex (0,0); focus (3,0)
10. Vertex (0,0); focus (–2,0)
11. Vertex (0,0) passing through (2,3) and axis is along x-axis.
12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

Ans

7.

Since, focus (6, 0) lies on x-axis and directrix
x = – 6 shows that parabola is in the form of
y2 = 4ax where a = 6.
Therefore, the equation of parabola is:
y2 = 4(6)x
y2 = 24 x

8.

Since, focus (0,–3) lies on y-axis and directrix
y = 3 shows that parabola is in the form of
x2 = – 4ay where a = 3.
Therefore, the equation of parabola is:
x2 = – 4(3)y
x2 = – 12y

9.

Since, vertex is (0, 0) and focus is (3, 0) which lies on x-axis of parabola.
Therefore, the equation of parabola will be in the form of y2 = 4ax.
Therefore,
y2 = 4(3)x
y2 = 12x is the required equation of parabola.

10.

Since, vertex is (0, 0) and focus is (–2, 0) which lies on x-axis of parabola.
Therefore, the equation of parabola will be in the form of y2 = – 4ax.
Therefore,
y2 = – 4(2)x
y2 = – 8x is the required equation of parabola.

11.

Since, parabola is along x-axis. So, its possible equation may be either y2 = 4ax or y2 = – 4ax.
Since, parabola is passing through the point (2, 3), which lies in first quadrant.
So, the equation is of the form y2 = 4ax.
The point (2, 3) lies on y2 = 4ax, then
(3)2 = 4a(2)
9 = 8a
a = 9/8
Therefore, the equation of the parabola is
y2 = 4(9/8)x
= (9/2)x
Or 2y2 = 9x, which is required equation.

12.

Since, parabola is along y-axis. So, its possible equation may be either x2 = 4ay or x2 = – 4ay.
Since, parabola is passing through the point (5, 2),
which lies in first quadrant. So, the equation is of the form x2 = 4ay.
The point (5, 2) lies on x2 = 4ay, then
(5)2 = 4a(2)
25 = 8a
a = 25/8
Therefore, the equation of the parabola is
x2 = 4(25/8)y
= (25/2)y
Or 2x2 = 25y, which is required equation.

Q.11 In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

1.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{36}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{=}\mathbf{1}$

2.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{=}\mathbf{1}$

3.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{=}\mathbf{1}$

4.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{100}}\mathbf{=}\mathbf{1}$

5.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{49}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{36}}\mathbf{=}\mathbf{1}$

6.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{100}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{400}}\mathbf{=}\mathbf{1}$

7. 36x2 + 4y2 = 144
8. 16x2 + y2 = 16
9. 4x2 + 9y2 = 36

Ans

1.

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{36}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{y}}^{\text{2}}}{16}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=36{\text{and b}}^{\text{2}}=16\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{4}\\ \text{\hspace{0.17em}c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{36-16}\\ \text{\hspace{0.17em}c}=±\text{\hspace{0.17em}}\sqrt{20}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(±\sqrt{20},0\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(±6,0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×6\\ \text{\hspace{0.17em}\hspace{0.17em}}=12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×4\\ \text{\hspace{0.17em}\hspace{0.17em}}=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{20}}{6}\\ \text{\hspace{0.17em}}\mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(4\right)}^{2}}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{16}{3}\end{array}$

2.
$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{25}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{4}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=25{\text{and b}}^{\text{2}}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}2}\\ \text{\hspace{0.17em}c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{25-4}\\ \text{\hspace{0.17em}c}=±\text{\hspace{0.17em}}\sqrt{21}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(0,±\text{\hspace{0.17em}}\sqrt{21}\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(0,±\text{\hspace{0.17em}}5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×5\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\end{array}$

$\begin{array}{l}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×2\\ \text{\hspace{0.17em}\hspace{0.17em}}=4\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{21}}{5}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(2\right)}^{2}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{8}{5}\end{array}$

3.

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{16}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{y}}^{\text{2}}}{9}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=16{\text{and b}}^{\text{2}}=9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}3}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{16-9}\\ \text{\hspace{0.17em} c}=±\text{\hspace{0.17em}}\sqrt{7}\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of foci}=\left(±\text{\hspace{0.17em}}\sqrt{7},0\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(±\text{\hspace{0.17em}}4,0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×4\\ \text{\hspace{0.17em}\hspace{0.17em}}=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×3\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{7}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(3\right)}^{2}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{9}{2}\end{array}$

4.

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{100}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{25}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=100{\text{and b}}^{\text{2}}=25\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}10\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}5}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{100-25}\\ \text{\hspace{0.17em} c}=±\text{\hspace{0.17em}}\sqrt{75}=±\text{\hspace{0.17em}}5\sqrt{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(0,±\text{\hspace{0.17em}}5\sqrt{3}\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(0,±\text{\hspace{0.17em}}10\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×10\\ \text{\hspace{0.17em}\hspace{0.17em}}=20\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×5\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{5\sqrt{3}}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}}{2}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(5\right)}^{2}}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}}=5\end{array}$

5.

$\text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{49}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}$ $\begin{array}{l}\text{of}\frac{{\text{y}}^{\text{2}}}{36}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=49{\text{and b}}^{\text{2}}=36\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}7\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}6}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{49-36}\\ \text{\hspace{0.17em} c}=±\text{\hspace{0.17em}}\sqrt{13}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(±\text{\hspace{0.17em}}\sqrt{13},0\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(±\text{\hspace{0.17em}}7,0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×7\\ \text{\hspace{0.17em}\hspace{0.17em}}=14\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×6\\ \text{\hspace{0.17em}\hspace{0.17em}}=12\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{13}}{7}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(6\right)}^{2}}{7}\end{array}$  $\mathrm{Length}\text{of latus rectum}=\frac{72}{7}$

6.

$\begin{array}{l}\text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{400}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{100}.\text{Comparing the given equation with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=400{\text{and b}}^{\text{2}}=100\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}20\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}10}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{400-100}\\ \text{\hspace{0.17em} c}=±\text{\hspace{0.17em}}\sqrt{300}=±\text{\hspace{0.17em}}10\sqrt{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(0,±\text{\hspace{0.17em}}10\sqrt{3}\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(0,±\text{\hspace{0.17em}}20\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×20\\ \text{\hspace{0.17em}\hspace{0.17em}}=40\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×10\\ \text{\hspace{0.17em}\hspace{0.17em}}=20\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{10\sqrt{3}}{20}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{3}}{2}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(10\right)}^{2}}{20}\\ \text{\hspace{0.17em}\hspace{0.17em}}=10\end{array}$

7.

$\begin{array}{l}\mathrm{The}\text{given equation of ellipse is:}\\ 36{\mathrm{x}}^{2}+4{\mathrm{y}}^{2}=144\\ \mathrm{Dividing}\text{both sides by 144, we get}\\ \frac{36{\mathrm{x}}^{2}}{144}+\frac{4{\mathrm{y}}^{2}}{144}=\frac{144}{144}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{4}+\frac{{\mathrm{y}}^{2}}{36}=1\\ \text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{36}\text{\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{4}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=36{\text{\hspace{0.17em}and \hspace{0.17em}b}}^{\text{2}}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}\hspace{0.17em}}6\text{\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}2}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{36-4}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}c}=±\text{\hspace{0.17em}}\sqrt{32}=±\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\sqrt{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(0,±\text{\hspace{0.17em}\hspace{0.17em}}4\text{\hspace{0.17em}}\sqrt{2}\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(0,±\text{\hspace{0.17em}\hspace{0.17em}}6\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×6\\ \text{\hspace{0.17em}\hspace{0.17em}}=12\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×2\\ \text{\hspace{0.17em}\hspace{0.17em}}=4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{4\text{\hspace{0.17em}}\sqrt{2}}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2\sqrt{2}}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(2\right)}^{2}}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{4}{3}\end{array}$

8.

$\begin{array}{l}\mathrm{The}\text{given equation of ellipse is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}16x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{16}\\ \mathrm{Dividing}\text{both sides by 144, we get}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{16{\mathrm{x}}^{2}}{16}+\frac{{\mathrm{y}}^{2}}{16}=\frac{16}{16}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{1}+\frac{{\mathrm{y}}^{2}}{16}=1\\ \text{Since},\text{the denominator of}\frac{{\text{y}}^{\text{2}}}{16}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{x}}^{\text{2}}}{1}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=16{\text{\hspace{0.17em}and \hspace{0.17em}b}}^{\text{2}}=1\\ ⇒\text{\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}\hspace{0.17em}}4\text{\hspace{0.17em}\hspace{0.17em}and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}1}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{16-1}\\ \text{\hspace{0.17em} c}=±\text{\hspace{0.17em}}\sqrt{15}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(0,±\text{\hspace{0.17em}\hspace{0.17em}}\sqrt{15}\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(0,±\text{\hspace{0.17em}\hspace{0.17em}}4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×4\\ \text{\hspace{0.17em}\hspace{0.17em}}=8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×1\\ \text{\hspace{0.17em}\hspace{0.17em}}=2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{15}}{4}\\ \text{\hspace{0.17em}}\mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{2{\left(1\right)}^{2}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{1}{2}\end{array}$

9.

$\begin{array}{l}\mathrm{The}\text{given equation of ellipse is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 4x}}^{\text{2}}+{\text{9y}}^{\text{2}}=\text{36}\\ \mathrm{Dividing}\text{both sides by 36, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{4{\mathrm{x}}^{2}}{\text{36}}+\frac{9{\mathrm{y}}^{2}}{\text{36}}=\frac{\text{36}}{\text{36}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{4}=1\\ \text{Since},\text{the denominator of}\frac{{\text{x}}^{\text{2}}}{9}\text{\hspace{0.17em}\hspace{0.17em}is greater than the denominator}\mathrm{}\\ \text{of}\frac{{\text{y}}^{\text{2}}}{4}.\text{Comparing the given equation}\mathrm{}\text{\hspace{0.17em}with}\\ \text{}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\mathrm{a}}^{2}=9{\text{\hspace{0.17em}\hspace{0.17em}and b}}^{\text{2}}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}3\text{and \hspace{0.17em}\hspace{0.17em}b}=±\text{\hspace{0.17em}2}\\ \text{\hspace{0.17em} c}=\sqrt{{\mathrm{a}}^{2}-{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{9-4}\\ \text{\hspace{0.17em} c}=±\text{\hspace{0.17em}}\sqrt{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Coordinates}\text{of foci}=\left(±\text{\hspace{0.17em}}\sqrt{5},0\right)\\ \mathrm{Coordinates}\text{of vertices}=\left(±\text{\hspace{0.17em}}3,0\right)\end{array}$ $\begin{array}{l}\mathrm{Length}\text{of major axis}=2\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×3\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\\ \mathrm{Length}\text{of minor axis}=2\mathrm{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2×2\\ \text{\hspace{0.17em}\hspace{0.17em}}=4\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{5}}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2{\left(2\right)}^{2}}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{8}{3}\end{array}$

Q.12 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Vertices}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{±}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{vertices are on x-axis, then equation of ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{where a is semi-major axis.}\\ \text{Given: \hspace{0.17em}\hspace{0.17em}a}=5\text{and c}=±\text{\hspace{0.17em}}4\\ \mathrm{Since},{\text{\hspace{0.17em}c}}^{\text{2}}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \therefore {\text{4}}^{\text{2}}={5}^{2}-{\mathrm{b}}^{2}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=25-16=9\\ \mathrm{Therefore},\text{equation of ellipse is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{25}+\frac{{\mathrm{y}}^{2}}{9}=1\end{array}$

Q.13 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Vertices}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{13}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{5}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{vertices are on y-axis, then equation of ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=13\text{and c}=±\text{\hspace{0.17em}}5\\ \mathrm{Since},{\text{\hspace{0.17em}c}}^{\text{2}}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \therefore {\text{5}}^{\text{2}}={13}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=169-25=144\\ \mathrm{Therefore},\text{equation of ellipse is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{144}+\frac{{\mathrm{y}}^{2}}{169}=1\end{array}$

Q.14 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Vertices}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{±}\mathbf{\text{\hspace{0.17em}}}\mathbf{6}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{vertices are on x-axis, then equation of ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=6\text{and c}=±\text{\hspace{0.17em}}4\\ \mathrm{Since},{\text{\hspace{0.17em}c}}^{\text{2}}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \therefore {\text{4}}^{\text{2}}={6}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=36-16=20\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{equation of ellipse is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{36}+\frac{{\mathrm{y}}^{2}}{20}=1\end{array}$

Q.15 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{major}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{±}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{minor}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{2}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Ends}\mathrm{of}\mathrm{major}\mathrm{axis}\text{\hspace{0.17em}}\left(±3,0\right),\mathrm{Ends}\mathrm{of}\mathrm{minor}\mathrm{axis}\left(0,±2\right)\\ \mathrm{Since},\text{vertices are on x-axis.Then equation of ellipse is}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=\text{3 and b}=2\\ \mathrm{Then},\text{equation of required ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{{3}^{2}}+\frac{{\mathrm{y}}^{2}}{{2}^{2}}=1\\ ⇒\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{4}=1\end{array}$

Q.16 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{major}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\sqrt{\mathbf{5}}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{Ends}\mathbf{}\mathbf{of}\mathbf{}\mathbf{minor}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Ends}\mathrm{of}\mathrm{major}\mathrm{axis}\text{\hspace{0.17em}}\left(0,±\sqrt{5}\right),\mathrm{Ends}\mathrm{of}\mathrm{minor}\mathrm{axis}\left(±1,0\right)\\ \mathrm{Since},\text{vertices are on y-axis.Then equation of ellipse is}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1,\text{where a is semi-major axis.}\\ \text{Given: a}=\sqrt{5}\text{and b}=1\\ \mathrm{Then},\text{equation of required ellipse is:}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{1}^{2}}+\frac{{\mathrm{y}}^{2}}{{\left(\sqrt{5}\right)}^{2}}=1\\ ⇒\frac{{\mathrm{x}}^{2}}{1}+\frac{{\mathrm{y}}^{2}}{5}=1\end{array}$

Q.17 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Length}\mathbf{}\mathbf{of}\mathbf{}\mathbf{major}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{26}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{\text{\hspace{0.17em}}}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{A}$

Ans

$\begin{array}{l}\mathrm{Since},\text{foci}\left(±5,0\right)\text{lies on x-axis.So, equation of ellipse is}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{where a is semi-major axis.}\\ \text{Given: Length of major axis}\left(2\mathrm{a}\right)=26\\ \text{\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{26}{2}=13\text{and c}=5\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}^{2}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{5}^{2}={\left(13\right)}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}25=169-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=169-25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=144\\ \mathrm{The}\text{equation of ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{169}+\frac{{\mathrm{y}}^{2}}{144}=1\end{array}$

Q.18 Find the equation for the ellipse that satisfies the given condition:A

$\mathbf{Length}\mathbf{}\mathbf{of}\mathbf{}\mathbf{minor}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{16}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{\text{\hspace{0.17em}}}\mathbf{6}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Length}\text{of minor axis}=16,\\ \mathrm{foci}=\left(0,±\text{\hspace{0.17em}}6\right)\\ \mathrm{Length}\text{of semi-minor axis}=\frac{16}{2}\end{array}$ $\begin{array}{l}\mathrm{b}=8\\ \mathrm{Since},\text{foci lies on y-axis, so equation of ellipse is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}^{2}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{6}^{2}={\mathrm{a}}^{2}-{8}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}36={\mathrm{a}}^{2}-64\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}=36+64\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{64}+\frac{{\mathrm{y}}^{2}}{100}=1,\text{which is required problem.}\end{array}$

Q.19 Find the equation for the ellipse that satisfies the given condition:

$\mathbf{Foci}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{\left(}\mathbf{±}\mathbf{\text{\hspace{0.17em}}}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{4}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Length}\text{of semi-major axis, a}=4,\\ \mathrm{foci}=\left(±\text{\hspace{0.17em}}3,0\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}^{2}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{3}^{2}={4}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=16-9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\\ \mathrm{Since},\text{foci is on x-axis. So, equation of ellipse is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \text{Putting value of a and b in equation of ellipse, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{16}+\frac{{\mathrm{y}}^{2}}{7}=1\end{array}$

Q.20 Find the equation for the ellipse that satisfies the given condition:
b = 3, c = 4, centre at the origin; foci on the x axis.

Ans

$\begin{array}{l}\mathrm{Since},\text{foci is on x-axis. So, equation of ellipse is}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Given}:\mathrm{b}=3\text{and c}=4\\ \mathrm{So},{\mathrm{c}}^{2}={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{4}^{2}={\mathrm{a}}^{2}-{3}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}=16+9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±5\\ \mathrm{Therefore},\text{the equation of ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{{5}^{2}}+\frac{{\mathrm{y}}^{2}}{{3}^{2}}=1\\ ⇒\frac{{\mathrm{x}}^{2}}{25}+\frac{{\mathrm{y}}^{2}}{9}=1\end{array}$

Q.21 Find the equation for the ellipse that satisfies the given condition:
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of ellipse when major axis lies on y-axis.}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}=1...\left(\mathrm{i}\right)\\ \mathrm{where}\text{a is semi-major axis.}\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}\left(3,\text{\hspace{0.17em}\hspace{0.17em}}2\right).\text{\hspace{0.17em}}\mathrm{So},\\ \frac{{3}^{2}}{{\mathrm{b}}^{2}}+\frac{{2}^{2}}{{\mathrm{a}}^{2}}=1\end{array}$ $\begin{array}{l}⇒\frac{9}{{\mathrm{b}}^{2}}+\frac{4}{{\mathrm{a}}^{2}}=1...\left(\mathrm{ii}\right)\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}\left(1,\text{\hspace{0.17em}\hspace{0.17em}}6\right).\text{\hspace{0.17em}}\mathrm{So},\\ \frac{{1}^{2}}{{\mathrm{b}}^{2}}+\frac{{6}^{2}}{{\mathrm{a}}^{2}}=1\\ ⇒\frac{1}{{\mathrm{b}}^{2}}+\frac{36}{{\mathrm{a}}^{2}}=1...\left(\mathrm{iii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from 9}×\text{equation}\left(\mathrm{iii}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{9}{{\mathrm{b}}^{2}}+\frac{324}{{\mathrm{a}}^{2}}=9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{±\frac{9}{{\mathrm{b}}^{2}}±\frac{4}{{\mathrm{a}}^{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{320}{{\mathrm{a}}^{2}}=8\\ ⇒{\mathrm{a}}^{2}=\frac{320}{8}=40\\ \mathrm{Putting}{\text{value of a}}^{\text{2}}\text{in equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{b}}^{2}}+\frac{36}{40}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{{\mathrm{b}}^{2}}=1-\frac{36}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4}{40}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=10\\ \mathrm{Putting}{\text{value of a}}^{\text{2}}{\text{and b}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{{\mathrm{x}}^{2}}{10}+\frac{{\mathrm{y}}^{2}}{40}=1\end{array}$

Q.22 Find the equation for the ellipse that satisfies the given condition:
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Ans

$\begin{array}{l}\mathrm{Equation}\text{of ellipse when major axis lies on y-axis.}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1...\left(\mathrm{i}\right)\\ \mathrm{where}\text{a is semi-major axis.}\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}\left(4,\text{\hspace{0.17em}\hspace{0.17em}}3\right).\text{\hspace{0.17em}}\mathrm{So},\\ \frac{{4}^{2}}{{\mathrm{a}}^{2}}+\frac{{3}^{2}}{{\mathrm{b}}^{2}}=1\\ ⇒\frac{16}{{\mathrm{a}}^{2}}+\frac{9}{{\mathrm{b}}^{2}}=1...\left(\mathrm{ii}\right)\\ \mathrm{Since},\text{ellipse}\left(\mathrm{i}\right)\text{passes through the point}\left(6,\text{\hspace{0.17em}\hspace{0.17em}}2\right).\text{\hspace{0.17em}}\mathrm{So},\\ \text{\hspace{0.17em}}\frac{{6}^{2}}{{\mathrm{a}}^{2}}+\frac{{2}^{2}}{{\mathrm{b}}^{2}}=1\\ ⇒\frac{36}{{\mathrm{a}}^{2}}+\frac{4}{{\mathrm{b}}^{2}}=1...\left(\mathrm{iii}\right)\\ \mathrm{Subtracting}\text{4}×\text{equation}\left(\mathrm{ii}\right)\text{from 9}×\text{equation}\left(\mathrm{iii}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{324}{{\mathrm{a}}^{2}}+\frac{36}{{\mathrm{b}}^{2}}=9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{±\frac{64}{{\mathrm{a}}^{2}}±\frac{36}{{\mathrm{b}}^{2}}=±4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{260}{{\mathrm{a}}^{2}}=5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}=\frac{260}{5}=52\end{array}$ $\begin{array}{l}\mathrm{Putting}{\text{value of a}}^{\text{2}}\text{in equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{36}{52}+\frac{4}{{\mathrm{b}}^{2}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{4}{{\mathrm{b}}^{2}}=1-\frac{36}{52}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{16}{52}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4}{13}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=13\\ \mathrm{Putting}{\text{value of a}}^{\text{2}}{\text{and b}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{{\mathrm{x}}^{2}}{52}+\frac{{\mathrm{y}}^{2}}{13}=1\end{array}$

Q.23 In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

1.

$\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{=}\mathbf{1}$

2.

$\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{27}}\mathbf{=}\mathbf{1}$

3. 9y2 – 4x2 = 36
4. 16 x2 – 9y2 = 576
5. 5y2 – 9x2 = 36
6. 49y2 – 16x2 = 784

Ans

1.

$\begin{array}{l}\mathrm{Comparing}\mathrm{}\frac{{\mathrm{x}}^{2}}{16}-\frac{{\mathrm{y}}^{2}}{9}=1\mathrm{with}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\mathrm{we}\mathrm{get}\\ {\mathrm{a}}^{\mathrm{2}}=16\mathrm{and}{\mathrm{b}}^{\mathrm{2}}=9\\ ⇒ \mathrm{a}=±4\mathrm{and}\mathrm{b}=±3\\ \mathrm{ } \mathrm{c}=±\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ \mathrm{ } =±\sqrt{16+9}\\ \mathrm{ } =±\sqrt{25}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}} \mathrm{ } \mathrm{c}=±5\\ \mathrm{Coordinates}\text{of foci are}\left(±5,0\right).\\ \mathrm{Coordinates}\text{of vertices are}\left(±4,0\right).\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}}=\frac{5}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ =\frac{2{\left(3\right)}^{2}}{4}\\ =\frac{9}{2}\end{array}$

2.

$\begin{array}{l}\mathrm{Comparing}\text{}\frac{{\mathrm{y}}^{2}}{9}-\frac{{\mathrm{x}}^{2}}{27}=1\text{with}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\text{a}}^{\text{2}}=9{\text{and b}}^{\text{2}}=27\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±3\text{and b}=±3\sqrt{3}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{9+27}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{36}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±6\\ \mathrm{Coordinates}\text{of foci are}\left(0,±6\right).\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of vertices are}\left(0,±3\right).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{6}{3}=2\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×27}{3}\\ \text{\hspace{0.17em}}=18\end{array}$

3.

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ {\text{9y}}^{\text{2}}-{\text{4x}}^{\text{2}}=\text{36}\\ ⇒\frac{{\text{9y}}^{\text{2}}}{36}-\frac{{\text{4x}}^{\text{2}}}{36}=\frac{\text{36}}{36}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{y}}^{\text{2}}}{4}-\frac{{\text{x}}^{\text{2}}}{9}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{}\frac{{\text{y}}^{\text{2}}}{4}-\frac{{\text{x}}^{\text{2}}}{9}=1\text{with}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\text{a}}^{\text{2}}=4{\text{and b}}^{\text{2}}=9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±2\text{and b}=±3\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{4+9}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{13}\\ \mathrm{Coordinates}\text{of foci are}\left(0,±\sqrt{13}\right).\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of vertices are}\left(0,±2\right).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{\sqrt{13}}{2}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×9}{2}\\ \text{\hspace{0.17em}}=9\end{array}$

4.

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}16{\mathrm{x}}^{2}-9{\mathrm{y}}^{2}=576\\ ⇒\frac{16{\mathrm{x}}^{2}}{576}-\frac{9{\mathrm{y}}^{2}}{576}=\frac{576}{576}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}}{36}-\frac{{\text{y}}^{\text{2}}}{64}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{}\frac{{\text{x}}^{\text{2}}}{36}-\frac{{\text{y}}^{\text{2}}}{64}=1\text{with}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\text{a}}^{\text{2}}=36{\text{and b}}^{\text{2}}=64\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}6\text{and b}=±\text{\hspace{0.17em}}8\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{36+64}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{100}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±10\\ \mathrm{Coordinates}\text{of foci are}\left(±10,0\right).\end{array}$ $\begin{array}{l}\mathrm{Coordinates}\text{of vertices are}\left(±\text{\hspace{0.17em}}6,0\right).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}}=\frac{10}{6}\\ \text{\hspace{0.17em}}=\frac{5}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}}=\frac{2{\left(8\right)}^{2}}{6}\\ \text{\hspace{0.17em}}=\frac{64}{3}\end{array}$

5.

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ \text{\hspace{0.17em}\hspace{0.17em}}5{\mathrm{y}}^{2}-9{\mathrm{x}}^{2}=36\\ ⇒\frac{{\text{5y}}^{\text{2}}}{36}-\frac{{\text{9x}}^{\text{2}}}{36}=\frac{\text{36}}{36}\\ ⇒\frac{{\text{y}}^{\text{2}}}{\left(\frac{36}{5}\right)}-\frac{{\text{x}}^{\text{2}}}{4}=1...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{}\frac{{\text{y}}^{\text{2}}}{\left(\frac{36}{5}\right)}-\frac{{\text{x}}^{\text{2}}}{4}=1\text{with}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\text{a}}^{\text{2}}=\frac{36}{5}{\text{and b}}^{\text{2}}=4\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\frac{6}{\sqrt{5}}\text{and b}=±2\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{\frac{36}{5}+4}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{\frac{36+20}{5}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{\frac{56}{5}}\text{\hspace{0.17em}\hspace{0.17em}}=±\frac{2\sqrt{14}}{\sqrt{5}}\\ \mathrm{Coordinates}\text{of foci are}\left(0,±\frac{2\sqrt{14}}{\sqrt{5}}\right).\\ \mathrm{Coordinates}\text{of vertices are}\left(0,±\frac{6}{\sqrt{5}}\right).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\frac{2\sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{14}}{3}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×{2}^{2}}{\left(\frac{6}{\sqrt{5}}\right)}\end{array}$  $\mathrm{Length}\text{of latus rectum}=\frac{4\sqrt{5}}{3}$

6.

$\begin{array}{l}\mathrm{The}\text{given equation of hyperbola is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 49y}}^{\text{2}}-{\text{16x}}^{\text{2}}=\text{784}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{49y}}^{\text{2}}}{\text{784}}-\frac{{\text{16x}}^{\text{2}}}{\text{784}}=\frac{\text{784}}{\text{784}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\text{y}}^{\text{2}}}{16}-\frac{{\text{x}}^{\text{2}}}{49}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{}\frac{{\text{y}}^{\text{2}}}{16}-\frac{{\text{x}}^{\text{2}}}{49}=1\text{with}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1,\text{we get}\\ {\text{a}}^{\text{2}}=16{\text{and b}}^{\text{2}}=49\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±4\text{and b}=±7\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{{\mathrm{a}}^{2}+{\mathrm{b}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=±\sqrt{16+49}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=±\sqrt{65}\\ \mathrm{Coordinates}\text{of foci are}\left(0,±\sqrt{65}\right).\\ \mathrm{Coordinates}\text{of vertices are}\left(0,±4\right).\\ \mathrm{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{65}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{2{\mathrm{b}}^{2}}{\mathrm{a}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{2×{7}^{2}}{4}\\ \mathrm{Length}\text{of latus rectum}=\frac{49}{2}\end{array}$

Q.24 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of the form}\\ \frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Since},\text{}\mathrm{vertices}\left(±2,0\right),\mathrm{foci}\left(±3,0\right)\\ \mathrm{So},\text{a}=\text{2 and c}=3.\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ ⇒\text{\hspace{0.17em}}{\mathrm{b}}^{2}={3}^{2}-{2}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9-4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=\sqrt{5}\\ \mathrm{Therefore},\text{equation of hyperbola is}\\ \frac{{\mathrm{x}}^{2}}{{2}^{2}}-\frac{{\mathrm{y}}^{2}}{{\left(\sqrt{5}\right)}^{2}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{4}-\frac{{\mathrm{y}}^{2}}{5}=1\end{array}$

Q.25 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{5}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{8}\mathbf{\right)}$

Ans

$\mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of the form}$ $\begin{array}{l}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Since},\text{}\mathrm{vertices}\left(0,±5\right),\mathrm{foci}\left(0,±8\right)\\ \mathrm{So},\text{a}=\text{5 and c}=8.\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ ⇒\text{\hspace{0.17em}}{\mathrm{b}}^{2}={8}^{2}-{5}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=64-25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=39\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=\sqrt{39}\\ \mathrm{Therefore},\text{equation of hyperbola is}\\ \frac{{\mathrm{y}}^{2}}{{5}^{2}}-\frac{{\mathrm{x}}^{2}}{{\left(\sqrt{39}\right)}^{2}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{y}}^{2}}{25}-\frac{{\mathrm{x}}^{2}}{39}=1\end{array}$

Q.26 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{3}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{foci}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{5}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of the form}\\ \frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Since},\text{}\mathrm{vertices}\left(0,±3\right),\mathrm{foci}\left(0,±5\right)\\ \mathrm{So},\text{a}=\text{3 and c}=5.\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}={5}^{2}-{3}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25-9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=±4\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{equation of hyperbola is}\\ \frac{{\mathrm{y}}^{2}}{{3}^{2}}-\frac{{\mathrm{x}}^{2}}{{4}^{2}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{y}}^{2}}{9}-\frac{{\mathrm{x}}^{2}}{16}=1\end{array}$

Q.27 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{transverse}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{8}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Foci}\left(±5,0\right),\mathrm{the}\mathrm{transverse}\mathrm{axis}\mathrm{is}\mathrm{of}\mathrm{length}8\\ \mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of}\\ \text{the form}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Here},\text{}\mathrm{c}=5\text{and a}=\frac{8}{2}=4\\ âˆµ{\mathrm{b}}^{2}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={5}^{2}-{4}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25-16\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=±3\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{4}^{2}}-\frac{{\mathrm{y}}^{2}}{{3}^{2}}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{16}-\frac{{\mathrm{y}}^{2}}{9}=1\end{array}$

Q.28 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\mathbf{13}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{conjugate}\mathbf{}\mathbf{axis}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{24}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Foci}\left(0,\text{\hspace{0.17em}}±13\right),\mathrm{the}\mathrm{conjugate}\mathrm{axis}\mathrm{is}\mathrm{of}\mathrm{length}24.\\ \mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of}\end{array}$ $\begin{array}{l}\text{the form}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=13\text{and b}=\frac{24}{2}=12\\ âˆµ{\mathrm{a}}^{2}={\mathrm{c}}^{2}-{\mathrm{b}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={13}^{2}-{\left(12\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=169-144\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=±\text{\hspace{0.17em}}5\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{y}}^{2}}{{5}^{2}}-\frac{{\mathrm{x}}^{2}}{{\left(12\right)}^{2}}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{y}}^{2}}{25}-\frac{{\mathrm{x}}^{2}}{144}=1\end{array}$

Q.29 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{3}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{the}\mathbf{}\mathbf{latus}\mathbf{}\mathbf{rectum}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{8}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Foci}\left(±3\sqrt{5},0\right),\mathrm{the}\mathrm{latus}\mathrm{rectum}\mathrm{is}\mathrm{of}\mathrm{length}8.\\ \mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of}\\ \text{the form\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Here},\text{}\mathrm{c}=3\sqrt{5}\text{and}\frac{{\text{2b}}^{\text{2}}}{\mathrm{a}}=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=4\mathrm{a}\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{a}={\left(3\sqrt{5}\right)}^{2}-{\mathrm{a}}^{2}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{a}=45-{\mathrm{a}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}+4\mathrm{a}-45=0\\ ⇒\text{\hspace{0.17em}}\left(\mathrm{a}+9\right)\left(\mathrm{a}-5\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=5,-9\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=5\left(\mathrm{Neglecting}\text{}-\text{9}\right)\\ \mathrm{and}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}^{\text{2}}=4×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{5}^{2}}-\frac{{\mathrm{y}}^{2}}{20}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{25}-\frac{{\mathrm{y}}^{2}}{20}=1\end{array}$

Q.30 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{the}\mathbf{}\mathbf{latus}\mathbf{}\mathbf{rectum}\mathbf{}\mathbf{is}\mathbf{}\mathbf{of}\mathbf{}\mathbf{length}\mathbf{}\mathbf{12}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Foci}\left(±4,0\right),\mathrm{the}\mathrm{latus}\mathrm{rectum}\mathrm{is}\mathrm{of}\mathrm{length}12.\\ \mathrm{Since},\text{foci is on x-axis, the equation of hyperbola is of}\\ \text{the form\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Here},\text{}\mathrm{c}=4\text{and}\frac{{\text{2b}}^{\text{2}}}{\mathrm{a}}=12\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=6\mathrm{a}\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{a}={4}^{2}-{\mathrm{a}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{a}=16-{\mathrm{a}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}+6\mathrm{a}-16=0\\ ⇒\text{\hspace{0.17em}}\left(\mathrm{a}+8\right)\left(\mathrm{a}-2\right)=0\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=2,-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=2\left(\mathrm{Neglecting}\text{}-\text{8}\right)\\ \mathrm{and}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b}}^{\text{2}}=6×2\left[âˆµ\text{\hspace{0.17em}}{\mathrm{b}}^{2}=6\mathrm{a}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{{2}^{2}}-\frac{{\mathrm{y}}^{2}}{12}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{4}-\frac{{\mathrm{y}}^{2}}{12}=1\end{array}$

Q.31 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Vertices}\mathbf{}\mathbf{\left(}\mathbf{±}\mathbf{7}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{,}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{e}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}$

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{Vertices}=\left(±7,0\right),\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{e}=\frac{4}{3}\\ \mathrm{Since},\text{vertices lie on x-axis, the equation of hyperbola is}\\ \text{of the form}\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Here},\text{a}=\text{7 and e}=\frac{4}{3}\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=\mathrm{ae}=7×\frac{4}{3}=\frac{28}{3}\\ \mathrm{and}{\text{\hspace{0.17em} b}}^{\text{2}}={\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ {\text{\hspace{0.17em} b}}^{\text{2}}={\left(\frac{28}{3}\right)}^{2}-{7}^{2}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{784}{9}-49\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{784-441}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{343}{9}\\ \mathrm{Therefore},\text{the equation of hyperbola is}\\ \frac{{\mathrm{x}}^{2}}{{7}^{2}}-\frac{{\mathrm{y}}^{2}}{\frac{343}{9}}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{x}}^{2}}{49}-\frac{9{\mathrm{y}}^{2}}{343}=1\end{array}$

Q.32 Find the equation of the hyperbola satisfying the below condition:

$\mathbf{Foci}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{±}\sqrt{\mathbf{10}}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{passing}\mathbf{}\mathbf{through}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{\right)}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Foci}\text{\hspace{0.17em}\hspace{0.17em}}\left(0,±\sqrt{10}\right),\mathrm{passing}\mathrm{through}\left(2,3\right).\\ \mathrm{Since},\text{foci is on y-axis, the equation of hyperbola is of}\\ \text{the form}\frac{{\mathrm{y}}^{2}}{{\mathrm{a}}^{2}}-\frac{{\mathrm{x}}^{2}}{{\mathrm{b}}^{2}}=1\\ \mathrm{Here},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=\sqrt{10}\\ âˆµ{\mathrm{c}}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}\\ ⇒10={\mathrm{a}}^{2}+{\mathrm{b}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{hyperbola passes through the point}\left(2,3\right).\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{3}^{2}}{{\mathrm{a}}^{2}}-\frac{{2}^{2}}{{\mathrm{b}}^{2}}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{9}{{\mathrm{a}}^{2}}-\frac{4}{{\mathrm{b}}^{2}}=1\end{array}$ $\begin{array}{l}\mathrm{Putting}{\text{b}}^{\text{2}}=10-{\mathrm{a}}^{2},\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{9}{{\mathrm{a}}^{2}}-\frac{4}{10-{\mathrm{a}}^{2}}=1\\ ⇒\frac{9\left(10-{\mathrm{a}}^{2}\right)-4{\mathrm{a}}^{2}}{{\mathrm{a}}^{2}\left(10-{\mathrm{a}}^{2}\right)}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}90-9{\mathrm{a}}^{2}-4{\mathrm{a}}^{2}=10{\mathrm{a}}^{2}-{\mathrm{a}}^{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{4}-23{\mathrm{a}}^{2}+90=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{a}}^{2}-18\right)\left({\mathrm{a}}^{2}-5\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}=18,5\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=10-{\mathrm{a}}^{2}\\ \text{}=10-5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\\ \mathrm{Neglecting}{\text{a}}^{\text{2}}=\text{18 because}{\mathrm{b}}^{2}=10-{\mathrm{a}}^{2}.\\ \mathrm{Therefore},\text{the equation of required hyperbola is}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{y}}^{2}}{5}-\frac{{\mathrm{x}}^{2}}{5}=1\end{array}$

Q.33 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Ans

Let point A(10, 5) lies on the parabolic reflector.
Equation of parabola is x2 = 4ay
Then, (10)2 = 4a(5)
100 = 20a
a = 100/20
= 5

Thus, the focus of the reflector is at the mid-point of the diameter.

Q.34 An arch is in the form of a parabola with its axis vertical. The arch is 10m high and 5 m wide at the base. How wide is it 2m from the vertex of the parabola?

Ans

Let point B(2.5, – 10) lies on parabolic arch. The equation of parabola is:
x2 = – 4ay …(i)
Putting x = 2.5 and y = – 10, we get
(2.5)2 = – 4a (–10)
6.25 = 40 a
4a = 0.625

Then, equation of parabolic arch is
x2 = – 0.625y …(ii)
Let the width of arch be 2b at height 2m from vertex. Then point A(b, – 2.5) lies on parabola,
Therefore,
b2 = – 0.625(– 2)
b2 = 1.25
b = 1.118
2b = 2 x 1.118
= 2.236 cm approx.
Thus, the width of arch at 2 m away from the vertex is 2.24 cm approx.

Q.35 The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Ans

Let road CF is hanging with cables. The length of the longest cable is 30 m and that of the smallest cable is 6 m. The smallest cable is 6 m long.

Road is hanging on parabolic cable.
Equation of parabola with vertex O(0, 0) is:
x2 = 4ay …(i)
Here, AB = 30 – 6 = 24 m and OB = 100/2 = 50 m
The coordinates of point A lying on parabolic cable is (50, 24). Let the coordinates of point P which is 18 m away from origin be (18, h).
Point A (50, 24) lies on parabola(i), then
(50)2 = 4a (24)
4a = 2500/24
= (625/6)
So, the equation of parabola is
x2 = (625/6)y …(ii)
Since, point P(18, h) lies on parabola (ii), then
(18)2 = (625/6)h
h = (324 x 6)/625
= 3.11 m
Therefore, length of cable which is 18 m away from the middle = 6 + 3.11
= 9.11 m.

Q.36 An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Ans

Since, width of semi-ellipse is 8 m and height of ellipse is 2 m. Coordinates of B and A are (4, 0) and (0, 2) respectively.
Here, BD = 1.5 m,
So, OD = OB – BD
= 4 – 1.5
= 2.5 m
Let height of CD be h m. Then, the coordinates of point C be (2.5, h).

$\begin{array}{l}\mathrm{Equation}\text{of ellipse is:}\\ \frac{{\mathrm{x}}^{2}}{{4}^{2}}+\frac{{\mathrm{y}}^{2}}{{2}^{2}}=1...\left(\mathrm{i}\right)\end{array}$ $\begin{array}{l}\text{Point C}\left(2.5,\mathrm{h}\right)\text{lies on ellipse}\left(\mathrm{i}\right),\text{then we have}\\ \frac{{\left(2.5\right)}^{2}}{{4}^{2}}+\frac{{\mathrm{h}}^{2}}{{2}^{2}}=1\\ ⇒\frac{6.25}{16}+\frac{{\mathrm{h}}^{2}}{4}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{h}}^{2}}{4}=1-\frac{6.25}{16}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{h}}^{2}}{4}=\frac{16-6.25}{16}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}=\frac{9.75}{16}×4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\sqrt{\frac{9.75}{16}×4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=1.56\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}\text{\hspace{0.17em}}\mathrm{approx}.\\ \mathrm{Thus},\text{the height of ellipse is 1.56 m\hspace{0.17em}\hspace{0.17em}}\mathrm{approx}\text{from a point}\\ \text{which is 1.5 meter away from on end.}\end{array}$

Q.37 A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Ans

Let AB be a rod of length 12 cm. A point P divides it into two parts i.e., AP and PB of length 3 m and (12 – 3) = 9 m respectively. PQ and PR are perpendiculars on X-axis and Y-axis respectively.

$\begin{array}{l}\text{InΔPQA},\text{sin}\mathrm{\theta }=\frac{\text{y}}{3}\text{}...\left(\text{i}\right)\\ \text{InΔPRB},\text{cos}\mathrm{\theta }=\frac{\text{x}}{9}\text{}...\left(\text{ii}\right)\\ \text{Squaringandaddingequation}\left(\text{i}\right)\text{andequation}\left(\text{ii}\right)\text{,}\\ \text{weget}\\ {\text{sin}}^{2}\mathrm{\theta }+{\text{cos}}^{2}\mathrm{\theta }={\left(\frac{\text{y}}{3}\text{}\right)}^{2}+{\left(\frac{\text{x}}{9}\right)}^{2}\\ 1=\frac{{\text{y}}^{2}}{9}+\frac{{\text{x}}^{2}}{81}\\ \text{or}\frac{{\text{x}}^{2}}{81}+\frac{{\text{y}}^{2}}{9}=1,\\ \text{whichistherequiredequationofpointPonanellipse.}\end{array}$

Q.38 Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.

Ans

Comparing x2 = 12y with x2 = 4ay, we get
4a = 12
a = 3
Coordinates of focus F = (0, 3)
Length of latus rectum = 4a
= 4(3)
= 12
Here, AF = FB = 12/2
= 6

$\begin{array}{l}\mathrm{Ar}\left(\mathrm{\Delta OAB}\right)=\frac{1}{2}|{\mathrm{x}}_{1}\left({\mathrm{y}}_{2}-{\mathrm{y}}_{3}\right)+{\mathrm{x}}_{2}\left({\mathrm{y}}_{3}-{\mathrm{y}}_{1}\right)+{\mathrm{x}}_{3}\left({\mathrm{y}}_{1}-{\mathrm{y}}_{2}\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}|0\left(3-3\right)+6\left(3-0\right)-6\left(0-3\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}|0\left(0\right)+6\left(3\right)-6\left(-3\right)|\\ \text{\hspace{0.17em}\hspace{0.17em}}=18\text{\hspace{0.17em}}\mathrm{square}\text{units}\end{array}$

Q.39 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Ans

Let A and B be the two flag posts at a distance of 8 m from each other. P(x, y) represents the position of a running man in such a way that:
PA + PB = 10 m

$\begin{array}{l}\mathrm{OA}=\mathrm{OB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8}{2}=4\\ \mathrm{So},\text{coordinates of A and B are}\left(4,0\right)\text{and}\left(-\text{\hspace{0.17em}}4,0\right)\text{respectively.}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PA}=\sqrt{{\left(\mathrm{x}-4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PB}=\sqrt{{\left(\mathrm{x}+4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}}\\ \mathrm{According}\text{to given condition:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PA}+\mathrm{PB}=\text{10}\\ ⇒\sqrt{{\left(\mathrm{x}-4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}}+\sqrt{{\left(\mathrm{x}+4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}}=10\end{array}$ $\begin{array}{l}⇒\text{}\sqrt{{\left(\text{x}-4\right)}^{2}+{\left(\text{y}-0\right)}^{2}}=10-\sqrt{{\left(\text{x}+4\right)}^{2}+{\left(\text{y}-0\right)}^{2}}\\ \mathrm{Squaring}\text{}\mathrm{both}\text{}\mathrm{sides},\text{}\mathrm{we}\text{}\mathrm{get}\\ {\text{x}}^{2}-8\text{x}+16+{\text{y}}^{2}=100+{\text{x}}^{2}+8\text{x}+16+{\text{y}}^{2}-20\sqrt{{\left(\text{x}+4\right)}^{2}+{\left(\text{y}-0\right)}^{2}}\\ \text{}0=100+16\text{x}-20\sqrt{{\left(\text{x}+4\right)}^{2}+{\left(\text{y}-0\right)}^{2}}\\ 20\sqrt{{\left(\text{x}+4\right)}^{2}+{\left(\text{y}-0\right)}^{2}}=100+16\text{x}\\ \mathrm{Again}\text{}\mathrm{squaring}\text{}\mathrm{both}\text{}\mathrm{sides},\text{}\mathrm{we}\text{}\mathrm{get}\\ 400\left({\text{x}}^{2}+8\text{x}+16+{\text{y}}^{2}\right)=10000+3200\text{x}+256{\text{x}}^{2}\\ \left(400-256\right){\text{x}}^{2}+400{\text{y}}^{2}=10000-6400\\ ⇒\text{}144{\text{x}}^{2}+400{\text{y}}^{2}=3600\\ \text{Dividing both sides by 3600, we get}\\ ⇒\text{}\frac{144}{3600}{\text{x}}^{2}+\frac{400}{3600}{\text{y}}^{2}=\frac{3600}{3600}\\ ⇒\text{}\frac{{\text{x}}^{2}}{25}+\frac{{\text{y}}^{2}}{9}=1\\ \mathrm{Which}\text{}\mathrm{is}\text{}\mathrm{the}\text{}\mathrm{required}\text{}\mathrm{equation}\text{}\mathrm{of}\text{}\mathrm{posts}\text{}\mathrm{traced}\text{}\mathrm{by}\text{}\mathrm{the}\text{}\mathrm{man}.\end{array}$

Q.40 An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Ans

Let each side of equilateral triangle be 2k units.

$\begin{array}{l}\mathrm{In}\text{}\mathrm{\Delta ACO},\text{by Pythagoras theorem,}\\ \text{\hspace{0.17em}\hspace{0.17em}OC}=\sqrt{{\mathrm{OA}}^{2}-{\mathrm{AC}}^{2}}\\ =\sqrt{{\left(2\mathrm{k}\right)}^{2}-{\mathrm{k}}^{2}}\\ =\sqrt{4{\mathrm{k}}^{2}-{\mathrm{k}}^{2}}\\ =\sqrt{3}\mathrm{k}\\ \mathrm{So},\text{coordinate of point A}=\left(\sqrt{3}\mathrm{k},\mathrm{k}\right)\\ {\text{Point A lies on parabola y}}^{\text{2}}=\text{4ax, then}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}k}}^{\text{2}}=\text{4a}\left(\sqrt{3}\mathrm{k}\right)\\ ⇒\mathrm{k}=\text{4a}\sqrt{3}\\ \mathrm{So},\text{each side of triangle}\\ =2\mathrm{k}=2\left(\text{4a}\sqrt{3}\right)\\ =8\sqrt{3}\mathrm{a}\\ \mathrm{Thus},\text{each side of equilateral triangle is}8\sqrt{3}\mathrm{a}.\end{array}$

Please register to view this section

1. How should I approach the Class 11 Mathematics Chapter 11?

Conic sections chapter is one of the complicated topics in Mathematics. One can approach the Class 11 Mathematics Chapter 11 with the proper understanding of the concepts covered in the NCERT textbook. You can also understand every topic through detailed solutions of the chapter related questions included inthe NCERT Solutions Class 11 Mathematics Chapter 11 which are available on the Extramarks’ website. Moreover, you can find additional  questions to practice in the NCERT solutions to enhance your performance. In this way, you can approach the chapter more smartly.

2. How can I make my Conic sections strong?

Regular practice helps better understand all the topics covered in the Conic sections. You can make it strong by practising problems of different difficulty levels and rectifying all mistakes. You can also find the solution to all the chapter related questions and additional questions in the NCERT Solutions Class 11 Mathematics Chapter 11 available on the Extramarks’ website.