If OB is the semi-minor axis of an ellipse, F_1 and F_2 are its foci, and the angle betwee…

Question

If $OB$ is the semi-minor axis of an ellipse, $F_1$ and $F_2$ are its foci, and the angle between $F_1B$ and $F_2B$ is a right angle, then the square of the eccentricity of the ellipse is:

Accepted Answer

$B=(0,b)$, $F_1=(-ae,0)$, $F_2=(ae,0)$. $F_1B\perp F_2B$:
$$\left(\frac{b}{ae}\right)\left(\frac{-b}{ae}\right)=-1\ \Rightarrow\ b^2=a^2e^2.$$
Using $b^2=a^2(1-e^2)$: $a^2(1-e^2)=a^2e^2\Rightarrow 1=2e^2\Rightarrow e^2=\tfrac12.$

Solutions · 1