# Class 11 Maths NCERT Solutions for Chapter 11 Conic Sections Exercise 11.4

### Conic Sections Exercise 11.4 Solutions

**1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola x ^{2}/16 - y^{2}/9 = 1 **

**Solution**

The given equation is

On comparing the equation with the standard equation of hyperbola i.e., x^{2} /a^{2} - y^{2} /b^{2} = 1, we obtain a = 4 and b = 3.

We know that a^{2} + b^{2} = c^{2}.

∴ c^{2} = 4^{2} + 3^{2} = 25

⇒ c = 5

Therefore,

The coordinates of the foci are (± 5, 0).

The coordinates of the vertices are (± 4, 0).

Eccentricity, e = c/a = 5/4

Length of latus rectum = 2b^{2}/a = (2×9)/4 = 9/2.

**2. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola y ^{2}/9 - x^{2}/27 = 1 **

**Solution**

The given equation is

On comparing this equation with the standard equation of hyperbola i.e., y^{2}/a^{2} - x^{2}/b^{2} = 1, we obtain a = 3 and b = √27.

We know that a^{2} + b^{2} = c^{2}.

∴ c^{2} = 3^{2} + (√27)^{2} = 9 + 27 = 36

⇒ c = 6

Therefore,

The coordinates of the foci are (0, ± 6).

The coordinates of the vertices are (0, ± 3).

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

Length of latus rectum = 2b^{2}/a = (2× 27)/3 = 18

**3. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y ^{2} – 4x^{2} = 36.**

**Solution**

The given equation is 9y^{2} – 4x^{2} = 36.

It can be written as

9y^{2} – 4x^{2} = 36

On comparing equation (1) with the standard equation of hyperbola i.e., y^{2}/a^{2} - x^{2}/b^{2} = 1, we obtain a = 2 and b = 3.

We know that a^{2} + b^{2} = c^{2}.

∴ c^{2} = 4 + 9 = 13

⇒ c = √13

Therefore,

The coordinates of the foci are (0, ± √13)

The coordinates of the vertices are (0, ± 2).

Eccentricity, e = c/a = √13/2

Length of latus rectum = 2b^{2}/a = (2× 9)/2 = 9

**4. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x ^{2} – 9y^{2} = 576 **

**Solution**

The given equation is 16x^{2} – 9y^{2} = 576.

It can be written as

16x^{2} – 9y^{2} = 576

On comparing equation (1) with the standard equation of hyperbola i.e., x^{2}/a^{2} - y^{2}/b^{2} = 1 , we obtain a = 6 and b = 8.

We know that a^{2} + b^{2} = c^{2}.

∴ c^{2} = 36 + 64 = 100

⇒ c = 10

Therefore,

The coordinates of the foci are (± 10, 0).

The coordinates of the vertices are ( ±6, 0).

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

Length of latus rectum =2b^{2}/a = (2 × 64)/6 = 64/3

**5. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y ^{2} – 9x^{2} = 36**

**Solution**

The given equation is 5y^{2} - 9x^{2} = 36.

On comparing equation (1) with the standard equation of hyperbola i.e., y^{2}/a^{2} - x^{2}/b^{2} = 1, we obtain a = 6/√5 and b = 2.

We know that a^{2} + b^{2} = c^{2}.

**6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y ^{2} – 16x^{2} = 784**

**Solution**

The given equation is 49y^{2} – 16x^{2} = 784.

It can be written as

49y^{2} – 16x^{2} = 784

On comparing equation (1) with the standard equation of hyperbola i.e., y^{2} /a^{2} - x^{2} /b^{2} = 1, we obtain a = 4 and b = 7

We know that a^{2} + b^{2} = c^{2}.

∴ c^{2} = 16 + 49 = 65

⇒ c = √65

Therefore,

The coordinates of the foci are (0, ± √65).

The coordinates of the vertices are (0, ± 4).

Eccentricity, e = c/a = √65/4

Length of latus rectum = 2b^{2}/a = (2× 49)/4 = 49/2

**7. Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0).**

**Solution**

Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} - y^{2}/b^{2} = 1.

Since the vertices are (±2, 0), a = 2.

Since the foci are (±3, 0), c = 3.

We know that a^{2} + b^{2} = c^{2}.

∴ 2^{2} + b^{2} = 3^{2}

b^{2} = 9 - 4 = 5

Thus, the equation of the hyperbola is x^{2}/4 - y^{2} /5 = 1.

**8. Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8).**

**Solution**

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form y^{2}/a^{2} - x^{2}/b^{2} = 1.

Since the vertices are (0, ±5), a = 5.

Since the foci are (0, ±8), c = 8.

We know that a^{2} + b^{2} = c^{2}.

∴ 5^{2} + b^{2} = 8^{2}

b^{2} = 64 - 25 = 39

Thus, the equation of the hyperbola is y^{2}/25 - x^{2}/39 = 1.

**9. Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5) . **

**Solution**

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form y^{2}/a^{2} - x^{2} /b^{2} = 1.

Since the vertices are (0, ±3), a = 3.

Since the foci are (0, ±5), c = 5.

We know that a^{2} + b^{2} = c^{2}.

∴3^{2} + b^{2} = 5^{2}⇒ b^{2} = 25 – 9 = 16

Thus, the equation of the hyperbola is y^{2}/9 - x^{2}/16 = 1.

**10. Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.**

**Solution**

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

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} - y^{2}/b^{2} = 1.

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8,

2a = 8

⇒ a = 4.

We know that a^{2} + b^{2} = c^{2}.

∴ 4^{2} + b^{2} = 5^{2}⇒ b^{2} = 25 – 16 = 9

Thus, the equation of the hyperbola is x^{2}/16 - y^{2}/9 = 1.

**11. Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.**

**Solution**

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

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form y^{2}/a^{2} - x^{2}/b^{2} = 1.

Since the foci are (0, ±13), c = 13.

Since the length of the conjugate axis is 24,

2b = 24

⇒ b = 12.

We know that a^{2} + b^{2} = c^{2}.

∴ a^{2} + 12^{2} = 13^{2}⇒ a^{2} = 169 – 144 = 25

Thus, the equation of the hyperbola is y^{2}/25 - x^{2}/144 = 1.

**12. Find the equation of the hyperbola satisfying the give conditions: Foci (± 3√5, 0), the latus rectum is of length 8.**

**Solution**

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

Here, the foci are on the x - axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} - y^{2}/b^{2} = 1.

Since the foci are (±3√5, 0), c = ± 3√5 .

Length of latus rectum = 8

⇒ 2b^{2} /a = 8

⇒ b^{2} = 4a

We know that a^{2} + b^{2} = c^{2} .

∴ a^{2} + 4a = 45

⇒ a^{2} + 4a – 45 = 0

⇒ a^{2} + 9a – 5a – 45 = 0

⇒ (a + 9) (a – 5) = 0

⇒⇒ a = –9, 5

Since a is non-negative, a = 5.

∴ b^{2} = 4a = 4 × 5 = 20

Thus, the equation of the hyperbola is x^{2}/25 - y^{2}/20 = 1.

**13. Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12 . **

**Solution**

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

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} - y^{2}/b^{2} = 1 .

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

⇒ 2b^{2}/a = 12

⇒ b^{2} = 6a

We know that a^{2} + b^{2} = c^{2}.

∴ a^{2} + 6a = 16

⇒ a^{2} + 6a – 16 = 0

⇒ a^{2} + 8a – 2a – 16 = 0

⇒ (a + 8) (a – 2) = 0

⇒ a = –8, 2

Since a is non-negative, a = 2.

∴ b^{2} = 6a = 6×2 = 12

Thus, the equation of the hyperbola is x^{2}/4 - y^{2}/12 = 1.

**14. Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), e = 4/3**

**Solution**

Vertices (±7, 0), e = 4/3

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form x^{2}/a^{2} - y^{2}/b^{2} = 1.

Since the vertices are (±7, 0), a = 7.

It is given that e = 4/3

Thus, the equation of the hyperbola is x^{2}/49 - 9y^{2}/343 = 1.

**15. Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±√10), passing through (2, 3) .**

**Solution**

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

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form y^{2}/a^{2} - x^{2}/b^{2} = 1.

Since the foci are (0, ± √10), c = √10 .

We know that a^{2} + b^{2} = c^{2}.

∴ a^{2} + b^{2} = 10

⇒ b^{2} = 10 – a^{2} **… (1)**

Since the hyperbola passes through point (2, 3),

9/a^{2} - 4/b^{2} = 1 **...(2) **

From equations (1) and (2), we obtain

⇒ 9(10 - a^{2} ) - 4a^{2} = a^{2} (10 - a^{2})

⇒ 90 - 9a^{2} - 4a^{2} = 10a^{2} - a^{4}

⇒ a^{4} - 23a^{2} + 90 = 0

⇒ a^{4} - 18a^{2} - 5a^{2} + 90 = 0

⇒ a^{2} (a^{2} - 18) - 5(a^{2} - 18) = 0

⇒ (a^{2} - 18)(a^{2} - 5) = 0

⇒ a^{2} = 18 or 5

In hyperbola, c > a, i.e., c^{2} > a^{2}

a^{2} = 5

b^{2} = 10 - a^{2} = 10 - 5 = 5

Thus, the equation of the hyperbola is y^{2}/5 - x^{2}/5 = 1.