# Hyperbola (S.N.Dey) | Part-6 | 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 Complete Long Answer Type questions of Hyperbola related problems of S N Dey mathematics class 11.

1. Find the centre, the length of latus rectum, the eccentricity, the co-ordinates of foci and the equations of the directrices of the hyperbola

Solution.

Comparing the given equation of hyperbola with the general form of hyperbola we get,

(i) The co-ordinates of centre of hyperbola :

(ii) The eccentricity

(iii) The co-ordinates of foci :

i.e., and

(iv) The equations of directrices are :

2. Show that the equation represents the equation of a hyperbola ; find the co-ordinates of its centre and foci and also the equation of its directrices.

Solution.

The equation of the hyperbola can be written as

Comparing the given equation of hyperbola with the general form of hyperbola we get,

(i) Centre :

(ii) Foci : i.e., and

The eccentricity

(iii) The equations of directrices are given by

3. Find (i) the centre (ii) the vertice (iii) the equations of the axes (iv) the lengths of axes (v) the eccentricities (vi) the lengths of latera recta (vii) the co-ordinates of foci (viii) the equations of the directrices of the following two hyperbolas :

Solution.

Comparing with we get,

(i) Centre of the hyperbola is

(ii) Co-ordinates of vertices :

(iii) The equation of transverse axis : and conjugate axis :

(iv)The length of transverse axis : and the length of conjugate axis :

(v) The eccentricity is given by

(vi) The length of latera recta :

(vii) The co-ordinates of foci :

(viii) The equations of directrices are given by

Solution (b)

Comparing with we get,

(i) Centre of the hyperbola is

(ii) Co-ordinates of vertices :

(iii) The equation of transverse axis : and conjugate axis :

(iv)The length of transverse axis : the length of conjugate axis :

(v) The eccentricity is given by

(vi) The length of latera recta :

(vii) The co-ordinates of foci :

(viii) The equations of directrices are given by

4. A point moves on a plane in such a manner that the difference of its distances from the pointsÂ  and is always constant and equal to . Show that the locus of the moving point is a rectangular hyperbola whose equation you are to determine.

Solution.

Let the co-ordinates of moving point be

So, by question,

Hence, by we can say that the locus of moving point is which is a rectangular hyperbola.

Note

5. A point moves on a plane so that its distance from the line Show that the locus of the moving point is a hyperbola andÂ  the equation of its locus is

Solution.

Let the co-ordinates of the moving point be

So, by question,

Hence, by we can say that the locus of the moving point represents a hyperbola and is given by

6. The hyperbola passes through the point of intersection of the lines and and the length of its latus rectum is Find the co-ordinates of its foci.

Solution.

The  given equations of straight lines are and

From and we get,

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

Since the given hyperbola passes through the point

Again, the length of the latus rectum of the hyperbola is

From and we get,

Hence, the co-ordinates of the foci are

7. The numerical value of the product of the perpendicular distances of a moving point from the lines and is Find the equation to the locus of the moving point.

Solution.

Comparing the given ellipse with the general equation of the ellipse we get,

The eccentricity of the given ellipse is

The co-ordinates of foci of the ellipse are

Let the co-ordinates of the moving point be

So, by question

Hence, the locus of the moving point is given by or,

8. An ellipse has the equation and and are its centre and focus in the usual notation. A hyperbola has a vertex at , has the point as its focus nearer to and has its latus rectum of equal length to that of Show that has the eccentricity and find its equation.

Solution.

The given ellipse

The eccentricity of is

The transverse axis of is along axis and the conjugate axis of H is along -axis. Let the eccentricity of H be the length of the transverse axis be and the length of the conjugate axis be

Focus of H is and the vertex    and so clearly, the ordinates of centre of is

By question, and

Again, the length of latus rectum of is = the length of latus rectum of .

For the hyperbola, the eccentricity is

hence, the equation of required hyperbola is

9. The foci of the ellipse coincide with the hyperbola. If for hyperbola, find the equation of the hyperbola.

Solution.

The given equation of ellipse is

Comparing with we get,

The eccentricity of is given by

The co-ordinates of foci of the ellipse

Let the equation of the hyperbola be

The eccentricity of the hyperbola . Since the  foci of the ellipse coincide with the hyperbola , so

For the hyperbola we get,

Hence, by ,  the equation of required hyperbola  is

Solution.

Clearly, the equation represents a hyperbola whose eccentricity is given by