question_answer

In an ellipse OB is a semi-minor axis, F and F' are its focus and angle FBF' is right angle. Then, eccentricity of the ellipse is

A)  \[\frac{1}{\sqrt{2}}\]

B)  \[\frac{1}{2}\]

C)  \[\frac{1}{4}\]

D)  \[\frac{1}{\sqrt{3}}\]

Correct Answer: A

Solution :

 \[\angle FBF'=90{}^\circ \] \[\Rightarrow \] \[\angle OBF'=45{}^\circ \] \[\Rightarrow \] \[ae=b\] and          \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \] \[{{e}^{2}}=1-\frac{{{e}^{2}}{{a}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \] \[{{e}^{2}}=1-{{e}^{2}}\] \[\Rightarrow \] \[2{{e}^{2}}=1\] \[\Rightarrow \] \[e=\frac{1}{\sqrt{2}}\]