NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise – Conic Sections (2025-26)

NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise covers real-world and advanced application-based problems on conic sections, including parabolas, ellipses, and their properties. This exercise integrates all four conic curves studied in the chapter and tests students' ability to apply geometric concepts to practical scenarios such as suspension bridges, reflectors, arches, and locus problems.

The Miscellaneous Exercise of Chapter 10 is a vital component of the CBSE Class 11 Maths syllabus 2025–26 and holds significant importance for board exam preparation. These application-based questions are regularly featured in competitive exams like JEE Main and JEE Advanced, making them essential for students aiming for top scores in both school and entrance examinations.

NCERT Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise – Conic Sections (2025-26)

Conic Sections

Medium

Q.

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. $\mathbf{centre}\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{)}\mathbf{and}\mathbf{ }\mathbf{radius}\mathbf{ }\frac{\mathbf{1}}{\mathbf{12}}$
4. $\mathbf{centre}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{and}\mathbf{ }\mathbf{radius}\mathbf{ }\sqrt{\mathbf{2}}$
5. $\mathbf{centre}\mathbf{(}\mathbf{-}\mathbf{a}\mathbf{,}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{radius}\mathbf{ }\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}$

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:
$\mathbf{Vertices}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{7}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{e}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}$

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Vertices}\mathbf{\quad }\mathbf{(}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}$$ 

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Vertices}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{8}\mathbf{)}$$ 

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $\mathbf{Vertices}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}$ 

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Foci}\mathbf{\quad }\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{the}\mathbf{\quad }\mathbf{transverse}\mathbf{\quad }\mathbf{axis}\mathbf{\quad }\mathbf{is}\mathbf{\quad }\mathbf{of}\mathbf{\quad }\mathbf{length}\mathbf{\quad }\mathbf{8}\mathbf{.}$$

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Foci}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{13}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{the}\mathbf{\quad }\mathbf{conjugate}\mathbf{\quad }\mathbf{axis}\mathbf{\quad }\mathbf{is}\mathbf{\quad }\mathbf{of}\mathbf{\quad }\mathbf{length}\mathbf{\quad }\mathbf{24}\mathbf{.}$$ 

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

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

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:

 $$\mathbf{Foci}\mathbf{\quad }\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{the}\mathbf{\quad }\mathbf{latus}\mathbf{\quad }\mathbf{rectum}\mathbf{\quad }\mathbf{is}\mathbf{\quad }\mathbf{of}\mathbf{\quad }\mathbf{length}\mathbf{\quad }\mathbf{12}\mathbf{.}$$ 

View Solution

Conic Sections

Medium

Q.

Find the equation of the hyperbola satisfying the below condition:
$\mathbf{Foci}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\sqrt{\mathbf{10}}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{passing}\mathbf{ }\mathbf{through}\mathbf{ }\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{)}\mathbf{.}$

View Solution

Conic Sections

Medium

Q.

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).

View Solution

Conic Sections

Easy

Q.

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

View Solution

Conic Sections

Medium

Q.

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?

View Solution

Conic Sections

Difficult

Q.

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.

View Solution

Conic Sections

Medium

Q.

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.

View Solution

Conic Sections

Medium

Q.

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.

View Solution

Conic Sections

Difficult

Q.

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.

View Solution

Conic Sections

Difficult

Q.

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.

View Solution

Conic Sections

Medium

Q.

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

View Solution

Conic Sections

Medium

Q.

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).

View Solution

Conic Sections

Medium

Q.

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

View Solution

Conic Sections

Medium

Q.

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.

View Solution

Conic Sections

Difficult

Q.

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.

View Solution

Conic Sections

Difficult

Q.

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.

View Solution

Conic Sections

Difficult

Q.

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

View Solution

Conic Sections

Difficult

Q.

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

View Solution

Conic Sections

Easy

Q.

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

View Solution

Conic Sections

Easy

Q.

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

View Solution

Conic Sections

Medium

Q.

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

View Solution

Conic Sections

Medium

Q.

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

View Solution

Conic Sections

Easy

Q.

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

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Vertices}\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
 $$\mathbf{Vertices}\mathbf{\text{\hspace{0.17em}}}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{13}\mathbf{)}\mathbf{,}\mathbf{\quad }\mathbf{foci}\mathbf{\quad }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{5}\mathbf{)}$$ 

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Vertices}\mathbf{(}\mathbf{\pm }\mathbf{6}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{4}\mathbf{,}\mathbf{0}\mathbf{)}$

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{major}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{minor}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{)}$

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{major}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\sqrt{\mathbf{5}}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{Ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{minor}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}$

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Length}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{major}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{26}\mathbf{,}\mathbf{ }\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{\pm }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{A}$

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:A
$\mathbf{Length}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{minor}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{16}\mathbf{,}\mathbf{ }\mathbf{foci}\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{6}\mathbf{)}$

View Solution

Conic Sections

Easy

Q.

Find the equation for the ellipse that satisfies the given condition:
$\mathbf{Foci}\mathbf{(}\mathbf{\pm }\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{a}\mathbf{=}\mathbf{4}$

View Solution

Conic Sections

Medium

Q.

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.

View Solution

Question 1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Origin is taken at the vertex, axis along positive x-axis. Equation: y² = 4ax. Passes through A(5, 10): 10² = 4 × a × 5 100 = 20a ⟹ a = 5

The focus = (a, 0) = (5, 0), which is the midpoint of the diameter. Therefore, the focus of the reflector is at the midpoint of the diameter.

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

Origin at vertex, axis along y-axis. Equation: x² = −4ay (opening downwards). Passes through (5/2, −10): (5/2)² = −4 × a × (−10) ⟹ a = 5/32

Equation: x² = −(5/8)y

At y = −2: x² = −(5/8)(−2) = 5/4 ⟹ x = √(5/4) m

Width = 2x = 2 × 1.118 ≈ 2.23 m

Therefore, the width of the arch 2 m from the vertex is approximately 2.23 m.

Question 3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway is 100 m long, the longest wire is 30 m and the shortest is 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Origin at lowest point of cable, axis along positive y-axis. Equation: x² = 4ay. AB = 30 m, OC = 6 m, BC = 50 m. Coordinates of A = (50, 24).

(50)² = 4a(24) ⟹ a = 625/24

Equation: 6x² = 625y

At x = 18: 6(18)² = 625y ⟹ y ≈ 3.11 m (= DE)

DF = DE + EF = 3.11 + 6 = 9.11 m

Therefore, the length of the supporting wire is approximately 9.11 m.

Question 4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the center. Find the height of the arch at a point 1.5 m from one end.

Semi-major axis a = 4, semi-minor axis b = 2. Equation: x²/16 + y²/4 = 1, y ≥ 0

At 1.5 m from one end: x = 4 − 1.5 = 2.5 m

(2.5)²/16 + y²/4 = 1 y² = 4(1 − 6.25/16) = 2.4375 ⟹ y ≈ 1.56 m

Therefore, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

Question 5. 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.

AP = 3 cm, PB = 9 cm. Let rod make angle θ with OX.

From ΔPBQ: cosθ = x/9 From ΔPRA: sinθ = y/3

Applying sin²θ + cos²θ = 1:

(y/3)² + (x/9)² = 1

Therefore, the equation of the locus is x²/81 + y²/9 = 1 (an ellipse).

Question 6. 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.

Comparing with x² = 4ay: a = 3. Focus S = (0, 3). At y = 3: x² = 36 ⟹ x = ±6 Vertices: O(0, 0), A(−6, 3), B(6, 3)

Area = (1/2)|0(3−3) + (−6)(3−0) + 6(0−3)| = (1/2)|−18 − 18| = (1/2)(36) = 18 sq. units

Therefore, the required area of the triangle is 18 sq. units.

Question 7. 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 path traced by the man.

PA + PB = 10 = 2a ⟹ a = 5 Distance between foci = 2c = 8 ⟹ c = 4 b² = a² − c² = 25 − 16 = 9 ⟹ b = 3

Therefore, the equation of the path traced by the man is x²/25 + y²/9 = 1.

Question 8. An equilateral triangle is inscribed in the parabola y² = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Let OC = k. Then A = (k, 2√(ak)) and B = (k, −2√(ak)). AB = 4√(ak)

Since OAB is equilateral, OA² = AB²: k² + 4ak = 16ak k² = 12ak ⟹ k = 12a

AB = 4√(a × 12a) = 4√(12a²) = 8√3 a

Therefore, the length of the side of the equilateral triangle inscribed in y² = 4ax is 8√3 a.

Frequently Asked Questions (FAQs)

Q1. What is the standard equation of a parabola opening to the right? The standard equation is y² = 4ax, where (a, 0) is the focus and x = −a is the directrix. The vertex is at the origin and the axis lies along the positive x-axis.

Q2. How do you find the equation of an ellipse given the sum of distances from two foci? If PA + PB = constant = 2a, the path is an ellipse. With foci at (±c, 0), use b² = a² − c² and write the equation as x²/a² + y²/b² = 1.

Q3. What is the latus rectum of a parabola? The latus rectum is the chord through the focus, perpendicular to the axis. For y² = 4ax, the length of the latus rectum is 4a.

Q4. How is the locus equation derived for a moving rod problem? For a rod of length L with ends on coordinate axes, if point P divides it at distances p and q from each end, then sinθ = y/p and cosθ = x/q. Applying sin²θ + cos²θ = 1 gives x²/q² + y²/p² = 1, which is an ellipse.

Q5. How many questions are in NCERT Class 11 Maths Chapter 10 Miscellaneous Exercise? The Miscellaneous Exercise of Chapter 10 – Conic Sections contains 8 questions covering parabola applications (reflectors, arches, suspension bridges), ellipse problems (racecourse, semi-ellipse arch), and locus problems involving conic sections.