NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Exercise 10.4 for 2026-27

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. x²/16 − y²/9 = 1

Solutions:

The given equation has a positive x² term, so the transverse axis is along the x-axis.

Comparing with the standard equation x²/a² − y²/b² = 1, we get a² = 16 and b² = 9, so a = 4 and b = 3.

Now, c = √(a² + b²) = √(16 + 9) = √25 = 5.

Coordinates of the foci = (± 5, 0)

Coordinates of the vertices = (± 4, 0)

Eccentricity, e = c/a = 5/4

Length of the latus rectum = 2b²/a = (2 × 9)/4 = 9/2

2. y²/9 − x²/27 = 1

Solutions:

Here the positive term is y², so the transverse axis is along the y-axis.

Comparing with y²/a² − x²/b² = 1, we get a² = 9 and b² = 27, so a = 3 and b = 3√3.

Now, c = √(a² + b²) = √(9 + 27) = √36 = 6.

Coordinates of the foci = (0, ± 6)

Coordinates of the vertices = (0, ± 3)

Eccentricity, e = c/a = 6/3 = 2

Length of the latus rectum = 2b²/a = (2 × 27)/3 = 18

3. 9y² − 4x² = 36

Solutions:

Dividing both sides by 36: y²/4 − x²/9 = 1.

The transverse axis is along the y-axis. Here a² = 4, b² = 9, so a = 2 and b = 3.

Now, c = √(a² + b²) = √(4 + 9) = √13.

Coordinates of the foci = (0, ± √13)

Coordinates of the vertices = (0, ± 2)

Eccentricity, e = c/a = √13/2

Length of the latus rectum = 2b²/a = (2 × 9)/2 = 9

4. 16x² − 9y² = 576

Solutions:

Dividing both sides by 576: x²/36 − y²/64 = 1.

The transverse axis is along the x-axis. Here a² = 36, b² = 64, so a = 6 and b = 8.

Now, c = √(a² + b²) = √(36 + 64) = √100 = 10.

Coordinates of the foci = (± 10, 0)

Coordinates of the vertices = (± 6, 0)

Eccentricity, e = c/a = 10/6 = 5/3

Length of the latus rectum = 2b²/a = (2 × 64)/6 = 64/3

5. 5y² − 9x² = 36

Solutions:

5. 5y² − 9x² = 36

Dividing both sides by 36: y²/(36/5) − x²/4 = 1.

The transverse axis is along the y-axis. Here a² = 36/5 and b² = 4, so a = 6/√5.

Now, c² = a² + b² = 36/5 + 4 = 56/5, so c = √(56/5) = 2√70/5.

Coordinates of the foci = (0, ± 2√70/5)

Coordinates of the vertices = (0, ± 6/√5), i.e., (0, ± 6√5/5)

Eccentricity, e = c/a = (2√70/5) ÷ (6/√5) = √14/3

Length of the latus rectum = 2b²/a = 8 ÷ (6/√5) = 8√5/6 = 4√5/3

6. 49y² − 16x² = 784

Solutions:

Dividing both sides by 784: y²/16 − x²/49 = 1.

The transverse axis is along the y-axis. Here a² = 16, b² = 49, so a = 4 and b = 7.

Now, c = √(a² + b²) = √(16 + 49) = √65.

Coordinates of the foci = (0, ± √65)

Coordinates of the vertices = (0, ± 4)

Eccentricity, e = c/a = √65/4

Length of the latus rectum = 2b²/a = (2 × 49)/4 = 49/2

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

7. Vertices (± 2, 0), foci (± 3, 0)

Solutions:

Since the vertices and foci lie on the x-axis, the equation is of the form x²/a² − y²/b² = 1.

From the vertices, a = 2; from the foci, c = 3.

Then b² = c² − a² = 9 − 4 = 5.

Required equation: x²/4 − y²/5 = 1

8. Vertices (0, ± 5), foci (0, ± 8)

Solutions:

Since the vertices and foci lie on the y-axis, the equation is of the form y²/a² − x²/b² = 1.

Here a = 5 and c = 8, so b² = c² − a² = 64 − 25 = 39.

Required equation: y²/25 − x²/39 = 1

9. Vertices (0, ± 3), foci (0, ± 5)

Solutions:

The foci lie on the y-axis, so the equation is of the form y²/a² − x²/b² = 1.

Here a = 3 and c = 5, so b² = c² − a² = 25 − 9 = 16.

Required equation: y²/9 − x²/16 = 1

10. Foci (± 5, 0), the transverse axis is of length 8

Solutions:

The foci lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1.

Length of the transverse axis = 2a = 8, so a = 4. Also c = 5.

Then b² = c² − a² = 25 − 16 = 9.

Required equation: x²/16 − y²/9 = 1

11. Foci (0, ± 13), the conjugate axis is of length 24

Solutions:

The foci lie on the y-axis, so the equation is of the form y²/a² − x²/b² = 1.

Length of the conjugate axis = 2b = 24, so b = 12. Also c = 13.

Then a² = c² − b² = 169 − 144 = 25.

Required equation: y²/25 − x²/144 = 1

12. Foci (± 3√5, 0), the latus rectum is of length 8

Solutions:

The foci lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1, with c = 3√5.

Length of the latus rectum = 2b²/a = 8, which gives b² = 4a.

Using c² = a² + b²: 45 = a² + 4a

a² + 4a − 45 = 0 ⟹ (a + 9)(a − 5) = 0 ⟹ a = 5 (rejecting a = −9, as a cannot be negative)

Then b² = 4 × 5 = 20.

Required equation: x²/25 − y²/20 = 1

13. Foci (± 4, 0), the latus rectum is of length 12

Solutions:

The foci lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1, with c = 4.

Latus rectum: 2b²/a = 12, so b² = 6a.

Using c² = a² + b²: 16 = a² + 6a

a² + 6a − 16 = 0 ⟹ (a + 8)(a − 2) = 0 ⟹ a = 2 (rejecting a = −8)

Then b² = 6 × 2 = 12.

Required equation: x²/4 − y²/12 = 1

14. Vertices (± 7, 0), e = 4/3

Solutions:

The vertices lie on the x-axis, so the equation is of the form x²/a² − y²/b² = 1, with a = 7.

Since e = c/a = 4/3, we get c = 7 × 4/3 = 28/3.

Then b² = c² − a² = 784/9 − 49 = (784 − 441)/9 = 343/9.

Required equation: x²/49 − y²/(343/9) = 1, i.e., x²/49 − 9y²/343 = 1

15. Foci (0, ± √10), passing through (2, 3)

Solutions:

The foci lie on the y-axis, so the equation is of the form y²/a² − x²/b² = 1, with c = √10, i.e., c² = 10, so b² = 10 − a².

Since the hyperbola passes through (2, 3):

9/a² − 4/(10 − a²) = 1

9(10 − a²) − 4a² = a²(10 − a²)

90 − 9a² − 4a² = 10a² − a⁴

a⁴ − 23a² + 90 = 0

Solving as a quadratic in a²: a² = [23 ± √(529 − 360)]/2 = (23 ± 13)/2, giving a² = 18 or a² = 5.

Since a² must be less than c² = 10, we take a² = 5, and then b² = 10 − 5 = 5.

Required equation: y²/5 − x²/5 = 1, i.e., y² − x² = 5

Key Formulas Used in Class 11 Maths Chapter 10 Exercise 10.4

Every question in this exercise comes down to three relations: c² = a² + b² (note the plus sign - the most common slip, since the ellipse uses a minus), eccentricity e = c/a which is always greater than 1 for a hyperbola, and latus rectum 2b²/a. Questions 12, 13, and 15 additionally form quadratic equations in a or a², so revise factorisation before attempting them. Practise similar problems from Vedantu's Chapter 10 Important Questions to build speed.

How to Practise NCERT Class 11 Maths Chapter 10 Conic Sections Exercise 10.4 for Full Marks with Vedantu’s Detailed Solutions?

Questions 1 to 6 of Exercise 10.4 are direct formula applications and ideal for one-mark and two-mark practice, while Questions 12 to 15 are the exam favourites — they combine the latus rectum or eccentricity condition with c² = a² + b² to form quadratics. In Question 15, remember to reject the root where a² exceeds c². After this exercise, move to the Miscellaneous Exercise on Chapter 10, where conic concepts appear in application form.

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Chapter-Specific NCERT Solutions for Class 11 Maths

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