A) \[{{e}_{1}}={{e}_{2}}\]
B) \[{{e}_{1}}>{{e}_{2}}\]
C) \[{{e}_{1}}<{{e}_{2}}\]
D) \[{{e}_{1}}+{{e}_{2}}=1\]
Correct Answer: B
Solution :
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 \[0<x<1\]
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