Question

If normal at $P(2,\frac{3\sqrt{3}}{2})$ meets the major axis of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ at $Q$ and $S,S^{\prime }$ are foci of given ellipse along positive and negative directions of axes, then the ratio $SQ:S^{\prime }Q$ is

• A. $\frac{7-\sqrt{8}}{7+\sqrt{8}}$
• B. $\frac{8-\sqrt{7}}{8+\sqrt{7}}$
• C. $\frac{7+\sqrt{8}}{7-\sqrt{8}}$
• D. $\frac{8+\sqrt{7}}{8-\sqrt{7}}$

Answer 1

The correct option is B $\frac{8-\sqrt{7}}{8+\sqrt{7}}$

Equation of the ellipse is
$\frac{x^2}{16}+\frac{y^2}{9}=1.$
$\Rightarrow e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

Hence, coordinates of foci are $S(\sqrt{7},0)$ and $S^{\prime }(-\sqrt{7},0).$


Normal at $P$ is a bisector of angle between $S^{\prime }P$ and $SP$
So, applying angular bisector theorem (angular bisector divides opposite side in the ration of its corresponding sides) in $△S^{\prime }PS$
i.e.,$\frac{SQ}{S^{\prime }Q}=\frac{SP}{S^{\prime }P}$
$\frac{SQ}{S^{\prime }Q}=\frac{a-e{x}_1}{a+e{x}_1}\Rightarrow \frac{SQ}{S^{\prime }Q}=\frac{4-\frac{\sqrt{7}}{2}}{4+\frac{\sqrt{7}}{2}}\Rightarrow \frac{SQ}{S^{\prime }Q}=\frac{8-\sqrt{7}}{8+\sqrt{7}}$