Question

If a ray of light incident along the line $3x+(5-4\sqrt{2})y=15$. gets reflected from the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ then its reflected ray goes along the line

• A. $x\sqrt{2}-y+5=0$
• B. $\sqrt{2}y-x+5=0$
• C. $\sqrt{2}y-x-5=0$
• D. $3x-y(4\sqrt{2}+5)+15=0$

Answer 1

Answer: (C)

$\sqrt{2}y-x-5=0$

Given equation of parabola is $\frac{x^2}{16}-\frac{y^2}{9}=1$

We have, for the given hyperbola

$9=16(e^2-1)$

$\Rightarrow e=\frac{5}{4}$

Since $(5,0)$ satisfies the equation of the line $3x+(5-4\sqrt{2})y=15,$ so the reflected must pass through $(-5,0)$ and

P is the point of intersection of curve and the ray

$P=(4\sqrt{2},3)$

$\therefore$ equation of $S^{\prime }P$ is $\sqrt{2}y=x+5.$