An ellipse has OB as semi minor axis, F and F | vedclass

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Question

(c) $\angle F'BF = 90^\circ $, $F'B \bot FB$

$i.e.$, slope of $(F'B)$ $	imes$ Slope of $(FB) = - 1$

==> $\frac{b}{{ae}} 	imes \

Vedclass · Accepted Answer

$\frac{1}{{\sqrt 2 }}$

Vedclass · Answer

(c) $\angle F'BF = 90^\circ $, $F'B \bot FB$

$i.e.$, slope of $(F'B)$ $	imes$ Slope of $(FB) = - 1$

==> $\frac{b}{{ae}} 	imes \frac{b}{{ - ae}} = - 1$, ${b^2} = {a^2}{e^2}$&hellip;..$(i)$

We know that $e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{{{a^2}{e^2}}}{{{a^2}}}} = \sqrt {1 - {e^2}} $

${e^2} = 1 - {e^2}$, $2{e^2} = 1$,

${e^2} = \frac{1}{2}$, $e = \frac{1}{{\sqrt 2 }}$.

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

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