• # question_answer In ellipse${{E}_{1}}$, the normal at one end of the latus rectum passes through one end of minor axis. In ellipse ${{E}_{2}}$, the latus rectum subtends a right angle at ${{E}_{1}},{{E}_{2}}$ respectively, then A) ${{e}_{1}}={{e}_{2}}$                   B) ${{e}_{1}}>{{e}_{2}}$ C) ${{e}_{1}}<{{e}_{2}}$                   D) ${{e}_{1}}+{{e}_{2}}=1$

Let${{E}_{1}}:\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ Equation of normal at $\left( a{{e}_{1}},\frac{{{b}^{2}}}{a} \right)$is$\frac{{{a}^{2}}x}{a{{e}_{1}}}-\frac{{{b}^{2}}y}{{{b}^{2}}}={{a}^{2}}{{e}^{2}}$it passes through (a,0), we get $e_{1}^{4}+e_{1}^{2}-1=0\Rightarrow {{e}_{1}}=\sqrt{\frac{\sqrt{5}-1}{2}};$ Let${{E}_{2}}:\frac{{{x}^{2}}}{{{c}^{2}}}+\frac{{{y}^{2}}}{{{d}^{2}}}=1$ Product of slopes of lines joining center of ellipse and ends of latus rectum is -1 $\frac{{{d}^{2}}}{c.c{{e}_{1}}}.\frac{-{{d}^{2}}}{c.c{{e}_{1}}}=-1\Rightarrow {{d}^{4}}={{c}^{4}}e_{1}^{2}$ $\Rightarrow$${{d}^{2}}={{c}^{2}}{{e}_{1}}$$\Rightarrow$${{c}^{2}}(1-e_{1}^{2})={{c}^{2}}{{e}_{1}}$ we get$e_{2}^{2}+{{e}_{2}}-1=0\Rightarrow {{e}_{2}}=\frac{\sqrt{5}-1}{2}$ $\therefore$${{e}_{1}}>{{e}_{2}}\sin ce\sqrt{x}>x$if $0<x<1$