Question

Let an incident ray $L_1=0$ gets reflected at point $A(-2,3)$ on hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ & passes through focus $S(2,0)$, then -

• A. equation of incident ray is $x+2=0$
• B. equation of reflected ray is $3x+4y=6$
• C. eccentricity, $e=2$
• D. length of latus rectum $=6$

Answer 1

Answer: (A), (B), (C), (D)

$S\equiv (2,0),S^{\prime }\equiv (-2,0)$
Using reflection property of hyperbola, S'A ${}^{\prime }$ is incident ray.
Equation of incident ray
$S^{\prime }A$ is $x=-2$
Equation of reflected ray
SP is $3x+4y=6$.
Now $2ae=4\Rightarrow ae=2$....(i)
Point $(-2,3)$ lies on hyperbola,
$\therefore \frac{4}{a^2}-\frac{9}{b^2}=1$
$\Rightarrow \frac{4}{a^2}-\frac{9}{4-a^2}=1$
on solving it we get $a=4$ (reject), $a=1$....(ii)
$\therefore$ Using (i) & (ii), we get $e=2$
length of latus rectum $=2a\left(e^2-1\right)=6$