Question

The area of the largest triangle that an incident ray and the corresponding reflected ray can enclose with the axis of the ellipse is equal to

• A. $4\sqrt{5}$
• B. $2\sqrt{5}$
• C. $\sqrt{5}$
• D. none of these

Answer 1

Answer: (B)

$2\sqrt{5}$

$\lambda x-y+2(1+\lambda )=0\Rightarrow \lambda (x+2)-(y-2)=0,$ passes through $(-2,2)$
$\mu x-y+2(1-\mu )=0\Rightarrow \mu (x-2)-(y-2)=0$, passes through $(2,2).$
Clearly these represent the foci of ellipse, so $2ae=4.$
The circle $x^2+y^2-4y-5=0\Rightarrow x^2+(y-2{)}^2=9$
represents auxiliary circle thus
$a^2=9\Rightarrow e=\frac{2}{3}$ and $b^2=5.$

Area of the largest triangle
$=\frac{1}{2}\times 2ae\times b=abe=2\sqrt{5}$