Question
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.
$$\frac{1}{{\sqrt 3 }}$$
Answer :
$$\frac{1}{{\sqrt 2 }}$$
Solution :
$$\eqalign{ & \because \,\angle FBF' = {90^ \circ } \Rightarrow F{B^2} + F'{B^2} = FF{'^2} \cr & \therefore {\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + {\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} = {\left( {2ae} \right)^2} \cr & \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2} \cr & \Rightarrow {e^2} = \frac{{{b^2}}}{{{a^2}}}......\left( {\text{i}} \right) \cr} $$
$$\eqalign{ & {\text{Also, }}{e^2} = 1 - \frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}\,\,\,\left( {{\text{By using equation}}\left( {\text{i}} \right)} \right) \cr & \Rightarrow 2{e^2} = 1 \cr & \Rightarrow e = \frac{1}{{\sqrt 2 }} \cr} $$
$$\eqalign{ & \because \,\angle FBF' = {90^ \circ } \Rightarrow F{B^2} + F'{B^2} = FF{'^2} \cr & \therefore {\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + {\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} = {\left( {2ae} \right)^2} \cr & \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2} \cr & \Rightarrow {e^2} = \frac{{{b^2}}}{{{a^2}}}......\left( {\text{i}} \right) \cr} $$
$$\eqalign{ & {\text{Also, }}{e^2} = 1 - \frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}\,\,\,\left( {{\text{By using equation}}\left( {\text{i}} \right)} \right) \cr & \Rightarrow 2{e^2} = 1 \cr & \Rightarrow e = \frac{1}{{\sqrt 2 }} \cr} $$