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.

• A. $3\sqrt{3}x-13y+15\sqrt{3}=0$
• B. $3\sqrt{3}x-13y-15\sqrt{3}=0$
• C. $3\sqrt{3}x+13y+15\sqrt{3}=0$
• D.

Answer 1

The correct option is A

$3\sqrt{3}x-13y+15\sqrt{3}=0$

Given hyperbola is $9x^2-16y^2=144$.

This equation can be rewritten as

$\frac{x^2}{16}-\frac{y^2}{9}=1$ - - - - - - (1)

Since x co-ordidate of p is 8.

Let y co-ordinate of p is $\alpha$.

$\because$ $(8,\alpha )$ lies on hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$

$\because$ $\frac{64}{16}-\frac{a^2}{9}=1$

$\Rightarrow {\alpha }^2=27$

$\Rightarrow \alpha =3\sqrt{3}$ $(\because$p lies in first quadrant)

Hence co-ordinate of point p is $(8,3\sqrt{3})$.

$\because$ Equation of reflected ray passes through

p$(8,3\sqrt{3})$ and other focus s'(-5,0)

$\therefore$ its equation is $y-3\sqrt{3}=\frac{0-3\sqrt{3}}{-5-8}(x-8)$

or $13y-39\sqrt{3}=3\sqrt{3}x-24\sqrt{3}$

or $3\sqrt{3}x-13y+15\sqrt{3}=0$