Question

The eccentricity of the ellipse whose latusrectum is equal to the distance between the foci, is ___________.

Answer 1

For an ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Given, latusrectum = Distance between foci
i.e $\frac{2b^2}{a}$ = 2ae
i.e 2b² = 2a²e
i.e $\frac{b^2}{a^2}$= e
$$\begin{gathered} \mathrm{Since}e=\sqrt{1-\frac{b^2}{a^2}} \\ \mathrm{i}.\mathrm{e}e=\sqrt{1-e} \\ \mathrm{i}.\mathrm{e}{e}^2=1-e \\ \mathrm{i}.\mathrm{e}{e}^2+e-1=0 \\ \mathrm{i}.\mathrm{e}e=\frac{-1\pm \sqrt{1+4}}{2} \\ e=\frac{\pm \sqrt{5}-1}{2} \\ \therefore \mathrm{eccentricity}\mathrm{is}\frac{\sqrt{5}-1}{2} \end{gathered}$$