Important Questions Class 11 Maths Chapter 11

Important Questions Class 11 Mathematics Chapter 11

Important Questions for CBSE Class 11 Mathematics Chapter 11 – Conic Sections

Conic Sections is Chapter 11 of the NCERT Book for Mathematics for Class 11. It is primarily concerned with various shapes, cones, and conic structures such as parabolas, hyperbolas, circles, axes, and the various subtopics associated with these. The following sections explain degenerate conic sections, standard equations of a parabola, Ellipse, Latus Rectum, and so on.

This is an important chapter that must be learned thoroughly through logical reasoning and practice. Students are advised to solve the Important Questions of this chapter, as each question is based on a different aspect and concept. Moreover, practising will improve problem-solving accuracy.

These topics are covered extensively in the CBSE examination. This chapter is also important in terms of competitive exams such as JEE Mains and JEE Advance. Students must master this chapter to excel in both school and competitive exams.

Extramarks Important Questions for Class 11 Mathematics Chapter 11 come with precise answers compiled by subject matter experts referring to the NCERT Books and past years’ question papers. Students can view the complete set of questions by clicking on the link provided here.

CBSE Class 11 Mathematics Chapter-11 Important Questions

(add important questions)

Access NCERT Solutions for Mathematics Class 11 – Chapter 11 – Conic Sections

Given below is a set of Chapter 11 Class 11 Maths Important Questions. 1-mark, 4-marks, and 6-mark answers and questions are discussed here.

Very Short Answer Questions: (1 Mark)

Q1. Find the length of the latus rectum of 2×2+3y2=18.

(a) 2 units

(b) 3 units

(c) 4 units

(d) None of these

A1. (c) 4 units

Q2. Find the length of the minor axis of x2+4y2=100.

(a) 10 units

(b) 12 units

(c) 14 units

(d) 8 units

A2. (a) 10 units

Q3. Find the equation of a circle with centre (b, a) & touching the x-axis.

(a) x2+y2−2bx+2ay+b2=0

(b) x2+y2+2bx−2ay+b2=0

(c) x2+y2−2bx−2ay+b2=0

(d) None of these

A3. (c) x2+y2−2bx−2ay+b2=0

Q4. Find the equations of the directrix & the axis of the parabola  ⇒3×2=8y 

(a) 3y−4=0, x=0

(b) 3x−2=0, X=0

(c) 3y−4x=0

(d) None of these

A4. (a) 3y−4=0, x=0

Long Answer Questions: (4 Marks)

Q1. Show that the equation x2+y2−6x+4y−36=0 represents a circle and find its centre & radius.

A1. It is of the form x2+y2+2gx+2Fy+c=0,

Where 2g=−6, 2f=4& c=−36

∴g=−3, f=2& c=−36

Thus, the centre of the circle is (−g,−f)=(3,−2)

The radius of the circle is g2+f2−c−−−−−−−−−√=9+4+36−−−−−−−−√

=7 units

Q2. Find the equation for the diameter of a circle drawn on the diagonal of a rectangle with sides x=6, x=3, y=3, and y=1.

A2. Let ABCD be the given rectangle and AD=x=−3, BC=x=6, AB=y=−1 & CD=y=−3.

Then A(−3,−1) and C(6,3).

The equation of the circle with AC as the diameter is:

(x+3)(x−6)+(y+1)(y−3)=0

⇒x2+y2−3x−2y−21=0

Q3. Show that the equation 6×2+6y2+24x−36y−18=0 represents a circle and find its centre & radius.

A3. 6×2+6y2+24x−36y−18=0

So, x2+y2+4x−6y+3=0

Where, 2g=4, 2f=−6 & c=3

∴g=2, f=−3 & c=3

Thus, the centre of the circle is (−g,−f)=(−2,3)

The radius of the circle =4+9+9−−−−−−−√=20−−√

=25–√ units

Q4. Find the equation for a circle whose endpoints for one of its diameters are A(2,3)& B(3,5).

A4. Let the endpoints of one of whose diameters are (x1,y1) and (x2,y2) be given by

(x−x1)(x−x2)+(y−y1)(y−y2)=0

So, x1=2, y1=−3 & x2=−3, y2=5.

∴ The required equation of the circle is (x−2)(x+3)+(y+3)(y−5)=0

⇒x2+y2+x−2y−21=0

Q5. Determine the focus and vertex coordinates, the equations of the directrix and the axis, and the length of the latus rectum of the parabola x2=8y.

A5. x2=−8y  & x2=−4ay

So, 4a=8

⇒a=2

So it is the downward parabola.

Foci are F(0,−a) i.e. F(0,−2).

Vertex is O(0,0).

So, y=a=2.

Its axis is y− axis, whose equation is given by x=0.

Length of latus rectum=4a units.

=4×2 units

=8 units

Very Long Answer Questions: (6 Marks)

Q1. A man running on a race track notices that the distance between the two flag posts from him is always 12 m, and the distance between the flag posts is 10 m. Determine the equation of the man’s path.

A1. An ellipse is the locus of a point that moves so that the sum of its distances from two fixed points remains constant.

As a result, the path is an ellipse.

Let the ellipse equation be x2/a2+y2/b2=1.

Where b2=a2(1−c2)

It is clear that 2a=12 & 2ae=10

⇒a=b and e=56

⇒b2=a2(1−e2)

⇒b2=36(1−2536)

⇒b2=11

So, the required equation is x2/36+y2/11=1.

Q2. Determine the axis lengths, vertices coordinates, and foci. the eccentricity and length of the hyperbola latus rectum 25×2−9y2=225.

A2. 25×2−9y2=225

⇒x2/9−y2/25=1

Now, a2=9 & b2=25

And c=a2+b2−−−−−−√

⇒c=9+25−−−−−√

⇒c=34−−√

(i) Length of transverse axis =2a=2×3=6 units

Length of conjugate axis =2b=2×5=10 units

(ii) The coordinates of vertices are A(−a,0) & B(a,0) i.e. A(−3,0) & B(3,0)

(iii) The coordinates of foci are F1(−c,0) & F2(c,0) i.e. F1(−34−−√,0) & F2(34−−√,0)

(iv) Eccentricity, e=ca=34−−√3

(v) Length of the latus rectum =2b2a=503 units

Q3. Find the equation of the curve formed by the set of all these points, the sum of whose distances from A(4,0,0) and B(4,0,0) is 10 units.

A3. Let P(x,y,z) be an arbitrary point on the given curve.

So, PA+PB=10

⇒(x−4)2+y2+z2−−−−−−−−−−−−−−√+(x+4)2+y2+z2−−−−−−−−−−−−−−√=10

=(x+4)2+y2+z2−−−−−−−−−−−−−−√=10−(x−4)2+y2+z2−−−−−−−−−−−−−−√

Squaring both sides:

⇒(x+4)2+y2+z2=100+(x−4)2+y2+z2−20(x−4)2+y2+z2−−−−−−−−−−−−−−√

⇒16x=100−20(x−4)2+y2+z2−−−−−−−−−−−−−−√

⇒5(x−4)2+y2+z2−−−−−−−−−−−−−−√=25−4x

⇒25[(x−4)2+y2+z2]=625+16×2−200x

⇒9×2+25y2+25z2−225=0

So, the required equation of the curve is 9×2+25×2+25z2−225=0.

Q4. An equilateral triangle is inscribed in the parabola y2=4ax so that one of the triangle’s angular points is at the parabola’s vertex. Determine the length of each triangle side.

A4. Assume OPQ is an equilateral triangle inscribed in the parabola y2=4ax with the vertex O(0,0), so ∠POM=∠QOM=30∘ .

Let OP=OQ=r.

And P=(rcos30∘,rsin30∘)

⇒P=(r3–√2,r2)

P lies on the parabola.

⇒r24=4a(r3–√2)

⇒r=8a3–√

Hence, the length of each side of the triangle is 8a3–√ units.