Question

Two rays of light coming along the line $x=4$ & $x=-2$ from the positive direction of $y$-axis & strikes a concave mirror whose intersection with $xy$ plane is $x^2=y$ at $A$ & $B$ respectively.The reflected ray passes through a fixed point $C$, then area of the triangle $ABC$ in square units is

• A. $\frac{81}{4}$
• B. $\frac{93}{4}$
• C. $\frac{91}{4}$
• D. $\frac{83}{4}$

Answer 1

The correct option is B $\frac{93}{4}$

Given parabola $x^2=y$
$\Rightarrow x^2=4\left(\frac{1}{4}\right)y$
As $x=4$
$\therefore y=16andx=-2$
$\therefore y=4$
$\therefore A=(4,16),B=(-2,4)$
Now after reflection both rays passes through the focus
$C(0,\frac{1}{4})$
Now,$A(4,16),B(-2,4),C(0,\frac{1}{4})$
$\therefore$ Area of $△ABC=\frac{1}{2}\left|\begin{array}{ccc}4& 16& 1\\ 2& 4& 1\\ 0& 1/4& 1\end{array}\right|=\frac{93}{4}$ square units
Hence the choice ( c) is correct answer