question_answer

An ellipse has OB as semi-minor axis, F and F' are its foci and the\[\angle FBF'\]is a right angle. Then, the eccentricity of the ellipse is

A) \[\frac{1}{\sqrt{3}}\]                                     

B) \[\frac{1}{4}\]

C) \[\frac{1}{2}\]                                   

D) \[\frac{1}{\sqrt{2}}\]

Correct Answer: D

Solution :

Since, \[\angle FBF'={{90}^{o}},\]then \[\angle OBF'={{45}^{o}}\]and \[\angle BF'O={{45}^{o}}\] \[\Rightarrow \]                               \[ae=b\] (\[\because \Delta \Beta OF'\]is an isosceles triangle)                 and        \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\]                 \[\Rightarrow \]               \[{{e}^{2}}=1-\frac{{{a}^{2}}{{e}^{2}}}{{{a}^{2}}}\Rightarrow {{e}^{2}}=1-{{e}^{2}}\]                 \[\Rightarrow \]               \[2{{e}^{2}}=1\Rightarrow e=\frac{1}{\sqrt{2}}\]                                 (\[\because e\]cannot be negative)