Question

A ray originating from the point (5,0) is incident on the hyperbola $9x^2-16y^2=144$ at the point P with abscissa 8. Find the equation of the reflected ray after first reflection and point P lying in first quadrant.

Answer 1

Given hyperbola is $9x^2-16y^2=144$ This equation can be

Rewritten as $\frac{x^2}{16}-\frac{y^2}{9}=1$

Since x co-ordinate of P is 8. Let y

Co-ordinate of P is a.

$\therefore$ (8,a) lies on (1)

$\therefore \frac{64}{16}-\frac{a^2}{9}=1\Rightarrow a^2=27\Rightarrow a=3\sqrt{3}$

Hence co-ordinate of point P is $(8,3\sqrt{3})\therefore$Equation of reflected ray passes through P $8,3\sqrt{3}$ and S(-5,0)

$\therefore$ Its equation is $y-3\sqrt{3}=0\frac{0-3\sqrt{3}}{-5-8}(x-8)or13y-39\sqrt{3}=3\sqrt{3}x-24\sqrt{3}or3\sqrt{3}x-13y+15\sqrt{3}=0$