Question

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

• A. $3\sqrt{3}x-13y+15\sqrt{3}=0$
• B. $3\sqrt{6}x-13y+15\sqrt{3}=0$
• C. $3\sqrt{3}x-13y+14\sqrt{3}=0$
• D. None of these

Answer 1

Answer: (A)

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

Let the point $P$ be $(8,k)$

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

$\Rightarrow 9{\left(8\right)}^2-16k^2=144\Rightarrow k=\pm 3\sqrt{3}$

As $P$ lies in the first quadrant, so its coordinates are $(8,3\sqrt{3})$

For the given hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$

We know $e^2=1+\frac{b^2}{a^2}$

$\Rightarrow e^2=1+\frac{9}{16}=\frac{25}{16}\Rightarrow e=\frac{5}{4}$

Focus is $(\pm ae,0)$.

Thus $S(5,0)$ and $S^{\prime }(-5,0)$

Reflected ray passes through $S^{\prime }$, so the equation of $P{S}^{\prime }$ is

$y-0=\frac{3\sqrt{3}-0}{8-(-5)}(x-(-5\left)\right)\Rightarrow 13y=3\sqrt{3}x+15\sqrt{3}\Rightarrow 3\sqrt{3}x-13y+15\sqrt{3}=0$

So, option A is correct.