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 [AIEEE 2005]

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

B)            \[\frac{1}{\sqrt{3}}\]

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

D)            \[\frac{1}{2}\]

Correct Answer: C

Solution :

           \[\angle {F}'BF=90{}^\circ \], \[{F}'B\bot FB\]                    i.e., slope of \[({F}'B)\] ´ Slope of \[({F}'B)=-1\]                    Þ \[\frac{b}{ae}\times \frac{b}{-ae}=-1\], \[{{b}^{2}}={{a}^{2}}{{e}^{2}}\]                        ?..(i)                               We know that \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{{{a}^{2}}{{e}^{2}}}{{{a}^{2}}}}=\sqrt{1-{{e}^{2}}}\]                    \[{{e}^{2}}=1-{{e}^{2}}\], \[2{{e}^{2}}=1\], \[{{e}^{2}}=\frac{1}{2}\], \[e=\frac{1}{\sqrt{2}}\].