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Q. A mirror in the first quadrant is in the shape of a hyperbola whose equation is $xy = 1$. A light source in the second quad rantem its a beam of light that hits the mirror at the point $(2, 1/2)$. If the reflected ray is parallel to the Y-axis, the slope of the incident beam is

KVPYKVPY 2015

Solution:

Equation of hyperbola $xy = 1$
image
$y=\frac{1}{x}$
$\frac{dy}{dx}=\frac{-1}{x^{2}}$
Slope of tangent at $\left(2, \frac{1}{2}\right)$
$\left(\frac{dy}{dx}\right)_{2, \frac{1}{2}}=\frac{-1}{2^{2}}=\frac{-1}{4}$
Slope of normal $=\frac{-1}{\frac{-1}{4}}=4$
Let the slope of beam of light = m
$\therefore \left|\frac{4-m}{1+4m}\right|=\left|\frac{\infty-4}{1-4\cdot\infty}\right|$
$\Rightarrow \frac{4-m}{1+4m}=\pm\frac{1}{4}$
$\Rightarrow m=\frac{15}{8}$