Question

A ray of light is coming along the line $y=b$ from the positive direction of x-axis & strikes a concave mirror whose intersection with the xy-plane is a parabola $y^2=4ax$. Let the equation of the reflected ray be $kabx+(4a^2-b^2)y-m{a}^{2}b=0$ Both $a$ & $b$ are positive. Find $k+m$ ?

Answer 1

Given parabola is $y^2=4ax$ ....(1)
Since, $y=b$ strikes the mirror.
$\Rightarrow x=\frac{b^2}{4a}$
So, the coordinates of P are $(\frac{b^2}{4a},b)$
Differentiating (1) w.r.t x
$2y\frac{dy}{dx}=4a$
Slope of tangent to (1) at P is $\frac{2a}{b}$
So, slope of normal is $-\frac{b}{2a}$
$\Rightarrow \tan ({180}^0-\alpha )=-\frac{b}{2a}$
$\Rightarrow \tan \alpha =\frac{b}{2a}$
Now, slope of reflected ray $=\tan ({180}^0-2\alpha )$
Slope of reflected ray $=-\tan 2\alpha$
$=-\frac{2\tan \alpha }{1-{\tan }^2\alpha }$
$=-\frac{4ab}{4a^2-b^2}$
Eqn of reflected ray is
$y-b=-\frac{4ab}{4a^2-b^2}(x-\frac{b^2}{4a})$
$\Rightarrow (y-b)(4a^2-b^2)=-4abx+b^3$
$\Rightarrow 4abx+\left(4a^2{b}^2\right)y+4a^{2}b=0$
Comparing with given equation of reflected ray
$k=4;m=4$
$\Rightarrow k+m=8$