## Question

### Solution

Correct option is Combined equation of asymptotes is  Also we know that the equation of the hyperbola differs from that of asymptotes by a constant.

Let the equation of the hyperbola be Since it passes through (1, –1) then   From (ii), equation of hyperbola is But we know that equation of conjugate hyperbola

= 2(Combined equation of asymptotes) – (Equation of hyperbola)   #### SIMILAR QUESTIONS

Q1

Find the locus of the foot of perpendicular from the centre upon any normal to the hyperbola .

Q2

Find the locus of the mid-points of the chords of the hyperbola which subtend a right angle at the origin.

Q3

Find the locus of the poles of normal chords of the hyperbola Q4

Find the condition for the lines Ax2 + 2Hxy + By2 = 0 to be conjugate diameters of .

Q5

Find the asymptotes of the hyperbola xy – 3y – 2x = 0.

Q6

A ray emanating from the point (5, 0) is incident on the hyperbola 9x2 – 16y2 = 144 at the point P with abscissa 8. Find the equation of the reflected ray after first reflection and point P lies in first quadrant.

Q7

The equations of the transverse and conjugate axes of a hyperbola are respectively 3x + 4y – 7 = 0, 4x – 3y + 8 = 0 and their respective lengths are 4 and 6. Find the equation of the hyperbola.

Q8

ABC are three points on the rectangular hyperbola xy = c2, find

1. The area of the triangle ABC

2. The area of the triangle formed by the tangents at AB and C

Q9

Find the coordinates of the foci and the equation of the directrices of the rectangular hyperbola xy = c2.

Q10

The vertices of the hyperbola 