Hyperbola (S.N.Dey) | Part-4 | Ex-6

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

4(i) For what value of the hyperbola will pass through the point Find its eccentricity and the length of latus rectum.

Solution.

Since the hyperbola passes through the point ,

So, the equation of the given hyperbola is

Comparing with the general form of hyperbola we get,

The eccentricity of the hyperbola is

The length of the latus rectum is

(ii) The hyperbola passes through the point of intersection of the lines and its eccentricity is ; show that its length of latus rectum is

Solution.

So, from and we get,

By we get,

So, the point of intersection of and is

Since the hyperbola   passes through the point

By we get,

Hence, the length of its latus rectum is

(iii) The hyperbola    passes through the point of intersection of the lines and and its latus rectum is Find and .

Solution.

From and we get,

From we get,

So, the point of intersection of the straight lines and is

Since the hyperbola   passes through the point

The length of latus rectum

By we get,

By we get,

(iv) The hyperbola    passes through the point and its eccentricity is find the length of its latus rectum.

Solution.

Since the hyperbola   passes through the point

By we get,

From we get,

the length of the latus rectum is

5. Find the equation of the hyperbola whose.

(i) eccentricity is focus is and the directrix is the line

Solution.

Let be any point on the hyperbola.

The distance of from the focus is given by

Again, the distance of from the directrix is

(ii) eccentricity is focus is and directrix is the line

Solution.

Let be any point on the hyperbola.

The distance of from the focus is given by

Again, the distance of from the directrix is

Hence, the equation represents the required equation of hyperbola.

(iii)eccentricity is focus is and the equation of directrix is

Solution.

Let be any point on the hyperbola.

The distance of from the focus is given by

Again, the distance of from the directrix is

Hence, the equation represents the required equation of hyperbola.

(iv)  eccentricity is focus is and the equation of directrix is

Let be any point on the hyperbola.

The distance of from the focus is given by

Again, the distance of from the directrix is

Hence, the equation represents the required equation of hyperbola.

(v)  eccentricity is focus is and the equation of directrix is

Solution.

Let be any point on the hyperbola.

The distance of from the focus is given by

Again, the distance of from the directrix is

Hence, the equation represents the required equation of hyperbola.

(vi) eccentricity is focus is and the equation of directrix is

Solution.

Let be any point on the hyperbola.

The distance of from the focus is given by

Again, the distance of from the directrix is

Hence, the equation represents the required equation of hyperbola.