## NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Ex 11.4

#### Page No 262:

#### Question 1:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

#### Answer:

The given equation is.

On comparing this equation with the standard equation of hyperbola i.e.,, we obtain

*a*= 4 and*b*= 3.
We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Therefore,

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

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

Length of latus rectum

#### Question 2:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

#### Answer:

The given equation is.

On comparing this equation with the standard equation of hyperbola i.e.,, we obtain

*a*= 3 and.
We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Therefore,

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

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

Length of latus rectum

#### Question 3:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9

*y*^{2}– 4*x*^{2}= 36#### Answer:

The given equation is 9

*y*^{2}– 4*x*^{2}= 36.
It can be written as

9

*y*^{2}– 4*x*^{2}= 36
On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain

*a*= 2 and*b*= 3.
We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Therefore,

The coordinates of the foci are.

The coordinates of the vertices are.

Length of latus rectum

#### Question 4:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16

*x*^{2}– 9*y*^{2}= 576#### Answer:

The given equation is 16

*x*^{2}– 9*y*^{2}= 576.
It can be written as

16

*x*^{2}– 9*y*^{2}= 576
On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain

*a*= 6 and*b*= 8.
We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Therefore,

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

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

Length of latus rectum

#### Question 5:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5

*y*^{2}– 9*x*^{2}= 36#### Answer:

The given equation is 5

*y*^{2}– 9*x*^{2}= 36.
On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain

*a*= and*b*= 2.
We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Therefore, the coordinates of the foci are.

The coordinates of the vertices are.

Length of latus rectum

#### Question 6:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49

*y*^{2}– 16*x*^{2}= 784#### Answer:

The given equation is 49

*y*^{2}– 16*x*^{2}= 784.
It can be written as 49

*y*^{2}– 16*x*^{2}= 784
On comparing equation (1) with the standard equation of hyperbola i.e.,, we obtain

*a*= 4 and*b*= 7.
We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Therefore,

The coordinates of the foci are.

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

Length of latus rectum

#### Question 7:

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

#### Answer:

Vertices (±2, 0), foci (±3, 0)

Here, the vertices are on the

*x*-axis.
Therefore, the equation of the hyperbola is of the form.

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}.
Thus, the equation of the hyperbola is.

#### Question 8:

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

#### Answer:

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the

*y*-axis.
Therefore, the equation of the hyperbola is of the form.

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}.
Thus, the equation of the hyperbola is.

#### Question 9:

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

#### Answer:

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the

*y*-axis.
Therefore, the equation of the hyperbola is of the form.

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.

#### Question 10:

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

#### Answer:

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.

Since the foci are (±5, 0),

*c*= 5.
Since the length of the transverse axis is 8, 2

*a*= 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.

#### Question 11:

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

#### Answer:

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.

Since the foci are (0, ±13),

*c*= 13.
Since the length of the conjugate axis is 24, 2

*b*= 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.

#### Question 12:

Find the equation of the hyperbola satisfying the give conditions: Foci, the latus rectum is of length 8.

#### Answer:

Foci, 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.

Since the foci are,

*c*=.
Length of latus rectum = 8

We know that

*a*^{2}+*b*^{2}=*c*^{2}.
∴

*a*^{2}+ 4*a*= 45
⇒

*a*^{2}+ 4*a*– 45 = 0
⇒

*a*^{2}+ 9*a*– 5*a*– 45 = 0
⇒ (

*a*+ 9) (*a*– 5) = 0
⇒

*a*= –9, 5
Since

*a*is non-negative,*a*= 5.
∴

*b*^{2}= 4*a*= 4 × 5 = 20
Thus, the equation of the hyperbola is.

#### Question 13:

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

#### Answer:

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.

Since the foci are (±4, 0),

*c*= 4.
Length of latus rectum = 12

We know that

*a*^{2}+*b*^{2}=*c*^{2}.
∴

*a*^{2}+ 6*a*= 16
⇒

*a*^{2}+ 6*a*– 16 = 0
⇒

*a*^{2}+ 8*a*– 2*a*– 16 = 0
⇒ (

*a*+ 8) (*a*– 2) = 0
⇒

*a*= –8, 2
Since

*a*is non-negative,*a*= 2.
∴

*b*^{2}= 6*a*= 6 × 2 = 12
Thus, the equation of the hyperbola is.

#### Question 14:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0),

#### Answer:

Vertices (±7, 0),

Here, the vertices are on the

*x*-axis.
Therefore, the equation of the hyperbola is of the form.

Since the vertices are (±7, 0),

*a*= 7.
It is given that

We know that

*a*^{2}+*b*^{2}=*c*^{2}.
Thus, the equation of the hyperbola is.

#### Question 15:

Find the equation of the hyperbola satisfying the give conditions: Foci, passing through (2, 3)

#### Answer:

Foci, passing through (2, 3)

Here, the foci are on the

*y*-axis.
Therefore, the equation of the hyperbola is of the form.

Since the foci are,

*c*=.
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),

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

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.

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