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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 eccentrieity $\frac{1}{3}$ and the distance of the nearer focus from this directrix is $\frac{8}{\sqrt{53}}$, then the equation of the other directrix can be :
$2 x-7 y-39=0$ or $2 x-7 y-7=0$
$11 x+7 y+8=0$ or $11 x+7 y-15=0$
$2 x-7 y+29=0$ or $2 x-7 y-7=0$
$11 x-7 y-8=0$ or $11 x+7 y+15=0$
Solution
Equation of reflected Ray
$y-1=\frac{2}{7}(x+2)$
$7 y-7=2 x+4$
$2 x-7 y+11=0$
Let the equation of other directrix is
$2 x-7 y+\lambda$
Distance of directrix from Focub
$\frac{a}{e}-a e=\frac{8}{\sqrt{53}}$
$3 a-\frac{a}{3}=\frac{8}{\sqrt{53}} \text { or } a=\frac{3}{\sqrt{53}}$
Distance between two directric $=\frac{2 \mathrm{a}}{\mathrm{e}}$
$=2 \times 3 \times \frac{3}{\sqrt{53}}=\frac{18}{\sqrt{53}}$
$\left|\frac{\lambda-11}{\sqrt{53}}\right|=\frac{18}{\sqrt{53}}$
$\lambda-11=18 \text { or }-18$
$\lambda=29 \text { or }-7$
$2 x-7 y-7=0 \text { or } 2 x-7 y+29=0$
