Question

A ray of light through $(2,1)$ is reflected at a point $P$ on the $y-$axis and then passes through the point $(5,3)$. If this reflected ray is the directrix of an ellipse with eccentricity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}},$ then the equation of the other directrix can be

• A. $2x-7y+29=0$ or $2x-7y-7=0$
• B. $11x+7y+8=0$ or $11x+7y-15=0$
• C. $2x-7y-39=0$ or $2x-7y-7=0$
• D. $11x-7y-8=0$ or $11x+7y+15=0$

Answer 1

The correct option is A $2x-7y+29=0$ or $2x-7y-7=0$

Image of $(2,1)$ w.r.t. $y-$axis is $(-2,1)$
$\therefore$ equation of reflected ray is
$y-1=\frac{3-1}{5+2}(x+2)$
$\Rightarrow 2x-7y+11=0\cdots \left(1\right)$
Let the equation of other directrix be $2x-7y+k=0$
Distance of directrix from focus
$\frac{a}{e}-ae=\frac{8}{\sqrt{53}}$
$\Rightarrow a(3-\frac{1}{3})=\frac{8}{\sqrt{53}}$
$\Rightarrow a=\frac{3}{\sqrt{53}}$
Now, distance between two directrices
$=\frac{2a}{e}=\frac{18}{\sqrt{53}}$
and $\left|\frac{k-11}{\sqrt{53}}\right|=\frac{18}{\sqrt{53}}$
$\Rightarrow |k-11|=18$
$\therefore k=29$ or $-7$
$\therefore$ Equation of directrix
$2x-7y+29=0$ or $2x-7y-7=0$