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# Chapter 11 Conic Sections Ex-11.4 Interview Questions Answers

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Question 1 : Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

The given equation is.

On comparing this equation with the standardequation of hyperbola i.e.,, we obtain a =4 and b = 3.

We know that a2 + b2 = c2.

Therefore,

The coordinates of thefoci are (±5, 0).

The coordinates of thevertices are (±4, 0).

Lengthof 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

The given equation is.

On comparing this equation with the standardequation of hyperbola i.e., , we obtain a =3 and.

We know that a2 + b2 = c2.

Therefore,

The coordinates of thefoci are (0, ±6).

The coordinates of thevertices are (0, ±3).

Lengthof latus rectum

Question 3 :

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola9y2 – 4x2 = 36

The given equation is 9y2 – 4x2 = 36.

It can be written as

9y2 – 4x2 = 36

On comparing equation (1) with the standardequation of hyperbola i.e.,, we obtain a =2 and b = 3.

We know that a2 + b2 = c2.

Therefore,

The coordinates of thefoci are.

The coordinates of thevertices are.

Lengthof latus rectum

Question 4 :

Find the coordinates of the foci and thevertices, the eccentricity, and the length of the latus rectum of the hyperbola16x2 – 9y2 = 576

The given equation is 16x2 – 9y2 = 576.

It can be written as

16x2 – 9y2 = 576

On comparing equation (1) with the standardequation of hyperbola i.e.,, we obtain a =6 and b = 8.

We know that a2 + b2 = c2.

Therefore,

The coordinates of thefoci are (±10, 0).

The coordinates of thevertices are (±6, 0).

Lengthof latus rectum

Question 5 :

Find the coordinates of the foci and thevertices, the eccentricity, and the length of the latus rectum of the hyperbola5y2 – 9x2 = 36

The given equation is 5y2 – 9x2 = 36.

On comparing equation (1) with the standardequation of hyperbola i.e.,, we obtain a =  and b = 2.

We know that a2 + b2 = c2.

Therefore, the coordinatesof the foci are.

The coordinates of thevertices are.

Lengthof latus rectum

Question 6 :

Find the coordinates of the foci and thevertices, the eccentricity, and the length of the latus rectum of the hyperbola49y2 – 16x2 = 784

The given equation is 49y2 – 16x2 = 784.

It can be written as 49y2 – 16x2 = 784

On comparing equation (1) with the standardequation of hyperbola i.e.,, we obtain a =4 and b = 7.

We know that a2 + b2 = c2.

Therefore,

The coordinates of thefoci are.

The coordinates of thevertices 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)

Vertices (±2, 0), foci(±3, 0)

Here, the vertices are on the x-axis.

Therefore, the equation ofthe hyperbola is of the form .

Since the vertices are (±2, 0), =2.

Since the foci are (±3, 0), c =3.

We know that a2 + b2 = c2.

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)

Vertices (0, ±5), foci (0,±8)

Here, the vertices are on the y-axis.

Therefore, the equation ofthe hyperbola is of the form.

Since the vertices are (0, ±5), =5.

Since the foci are (0, ±8), c =8.

We know that a2 + b2 = c2.

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)

Vertices (0, ±3), foci (0,±5)

Here, the vertices are on the y-axis.

Therefore, the equation ofthe hyperbola is of the form.

Since the vertices are (0, ±3), =3.

Since the foci are (0, ±5), c =5.

We know that a2 + b2 = c2.

32 + b2 = 52

b2 = 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.

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

Here, the foci are on the x-axis.

Therefore, the equation ofthe hyperbola is of the form.

Since the foci are (±5, 0), c =5.

Since the length of the transverse axis is8, 2a = 8  a = 4.

We know that a2 + b2 = c2.

42 + b2 = 52

b2 = 25 – 16 = 9

Thus,the equation of the hyperbola is.

krishan