question_answer

If OB is the semi-minor axis of an ellipse, \[{{F}_{1}}\] and \[{{F}_{2}}\] are its foci and the angle between \[{{F}_{1}}B\]and \[{{F}_{2}}B\] is a right angle, then the square of the eccentricity of the ellipse is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

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 \[{{F}_{2}}B=-1\] \[\left( \frac{0-b}{-ae-0} \right)\times \left( \frac{0-b}{ae-0} \right)=-1\] \[\left( \frac{b}{ae} \right)\times \left( \frac{-b}{ae} \right)=-1\] \[{{b}^{2}}={{a}^{2}}{{e}^{2}}\] \[{{e}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\]                 \[\left\{ \because {{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}} \right\}\] \[1-\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{{{b}^{2}}}{{{a}^{2}}}\]           \[1=2\frac{{{b}^{2}}}{{{a}^{2}}}\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{1}{2}\] \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}=1-\frac{1}{2}=\frac{1}{2}\]          \[{{e}^{2}}=\frac{1}{2}\]