A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: A
Solution :
[a] \[\because \angle FBF'=90{}^\circ \Rightarrow F{{B}^{2}}+F'{{B}^{2}}=FF{{'}^{2}}\] \[\therefore {{\left( \sqrt{{{a}^{2}}{{e}^{2}}+{{b}^{2}}} \right)}^{2}}+{{\left( \sqrt{{{a}^{2}}{{e}^{2}}+{{b}^{2}}} \right)}^{2}}={{(2ae)}^{2}}\] \[\Rightarrow 2({{a}^{2}}{{e}^{2}}+{{b}^{2}})=4{{a}^{2}}{{e}^{2}}\Rightarrow {{e}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\] ?. (i) Also, \[{{e}^{2}}=1-{{b}^{2}}/{{a}^{2}}=1-{{e}^{2}}\] (By using equation (i)) \[\Rightarrow 2{{e}^{2}}=1\Rightarrow e=\frac{1}{\sqrt{2}}.\]You need to login to perform this action.
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