# Hyperbola (S.N.Dey) |Part-3 | Ex-6

###### In the previous article , we have solved few VSA type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few Short Answer Type questions of Hyperbola related problems of s n dey mathematics class 11.

1. Find (i) the length of axes (ii) length of latus rectum (iii) co-ordinates  of vertices (iv) eccentricity (v) co-ordinates of foci and (vi) equations of the directrices of each of the following hyperbolas :

Solution (a)

Comparing with the general equation of hyperbola we get,

(i) The length of the transverse axis and the length of the conjugate axis

(ii) The length of latus rectum

(iii) The co-ordinates of vertices:

(iv) Eccentricity

(v) The co-ordinates of the foci :

(vi) Equations of directrices are

Solution(b)

Comparing with the general equation of hyperbola we get,

(i) The length of the transverse axis and the length of the conjugate axis

(ii) The length of latus rectum

(iii) The co-ordinates of vertices:

(iv) Eccentricity

(v) The co-ordinates of the foci :

(vi) Equations of directrices are

2. Find the length of the transverse and conjugate axes of the hyperbola Write down the equation of the hyperbola conjugate to it and find the eccentricities of both the hyperbolas.

Solution.

Comparing with the general equation of hyperbola we get,

So, the length of the transverse axis and the length of the conjugate axis

2nd Part :

The equation of the conjugate hyperbola is

3rd Part :

The eccentricity of the hyperbola is

The eccentricity of the conjugate hyperbola is

3. Find the equation of the hyperbola, whose axes are axes of co-ordinates and

(i) length of transverse and conjugate axes are and respectively.

Solution.

Since the transverse axis is along -axis and conjugate axis is along -axis, so the equation of the hyperbola can be written in the form of

So, by question,

Hence, the equation of the hyperbola is

(ii) lengths of conjugate axis and latus rectum are and respectively.

Solution.

By question, the length of conjugate axis and the length of latus rectum

Hence, the equation of the hyperbola is

(iii)which passes through the points and

Solution.

The hyperbola passes through the points and

So, from and we get,

So, by we get,

Hence, the required equation of the hyperbola is

(iv) distance between the foci is and length of conjugate axis is

Solution.

The length of conjugate axis

The distance between the foci

Hence, by and we get the required equation of hyperbola which is

(v) co-ordinates of foci are and the length of latus rectum is

Solution.

The co-ordinates of foci

The length of the latus rectum

From we get,

Hence, the equation of the hyperbola is

(vi)distances between the foci and directrices are and respectively.

Solution.

The distance between the foci

Again, the distance between two directrices

So, by , we get

Again, by we get,

So, the equation of the hyperbola is

(vii) eccentricity is and the sum of squares of the lengths of axes is

Solution.

By question, we have

Putting the value of in we get,

Hence, the equation of the hyperbola is

(viii)transverse axis is and the vertex bisects the line segment joining the centre and focus.

Solution.

Since the vertex bisects the line segment joining the centre and the focus,

the distance of the vertex from the centre

(the distance of the focus from the centre)

Hence, the equation of the hyperbola is

(ix) which passes through the point and whose eccentricity is .

Solution.

The hyperbola passes through the point

By we get,

So, using we get,

Hence, the equation of the hyperbola is given by

(x) eccentricity is and the co-ordinates of one focus are

Solution.

Eccentricity is and the co-ordinates of one focus are

The equation of the hyperbola is

(xi) eccentricity is and the length of semi latus rectum is

Solution.

It is given that the eccentricity is and the length of semi-latus rectum is

Hence, the equation of the hyperbola is

(xii)co-ordinates of foci are and the eccentricity is

Solution.

The co-ordinates of foci and the eccentricity is

By and we get,

Hence, the equation of the hyperbola is