question_answer

An ellipse has OB as semi minor axis, F and F? its foci and the angle FBF? is a right angle. Then the 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 :

[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)) \[\Rightarrow 2{{e}^{2}}=1\Rightarrow e=\frac{1}{\sqrt{2}}.\]