A) (a)\[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{2\sqrt{2}}\]
D) \[\frac{1}{4}\]
Correct Answer: A
Solution :
Let\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]be the equation of ellipse. Given that \[{{F}_{1}}B\]and \[{{F}_{2}}B\] are perpendicular to each other. Slope of \[{{F}_{1}}B\times \] slope of \[{{F}_{2}}B=-1\] \[\left( \frac{0-b}{-ae-0} \right)\times \left( \frac{0-b}{ae-0} \right)=-1\] \[\left( \frac{b}{ae} \right)\times \left( \frac{-b}{ae} \right)=-1\] \[{{b}^{2}}={{a}^{2}}{{e}^{2}}\] \[{{e}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\left\{ \because {{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}} \right\}\] \[1-\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{{{b}^{2}}}{{{a}^{2}}}\] \[1=2\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{1}{2}\] \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}=1-\frac{1}{2}=\frac{1}{2}\] \[{{e}^{2}}=\frac{1}{2}\]You need to login to perform this action.
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