Question 134 of 183MediumMCQ
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: Solutions · 1
Official solution
as \[\angle FBF' = {90^ \circ }\]
\[ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}\]
\[\therefore\] \[{\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}\]
\[ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}\]
\[ \Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}\]
Also e2=1−b2/a2=1−e2
\[ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}\] Answer the question to unlock