Question

If the eccentricity of an ellipse is $\frac{5}{8}$ and the distance between its foci is 10, then the length of the latusrectum is ___________.

Answer 1

For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
eccentricity = $\frac{5}{8}$
Distance between foci is 10 (given)
i.e 2ae = 10
i.e ae = 5
i.e a$\times \frac{5}{8}$ = 5
i.e a = 8
Length of latus rectum is $\frac{2b^2}{a}...\left(1\right)$
Since $e^2=1-{\left(\frac{b}{a}\right)}^2$
i.e $b^2=a^2\left(1-e^2\right)$
$$\begin{gathered} \therefore \mathrm{Length}\mathrm{of}\mathrm{latus}\mathrm{rectum}\mathrm{is}2\left(a^2\left(1-e^2\right)\right)\frac{1}{a} \\ =2a\left(1-e^2\right) \\ =2\times 8\left(1-\frac{25}{64}\right) \\ =16\left(\frac{64-25}{64}\right) \\ \mathrm{Hence},\mathrm{Length}\mathrm{of}\mathrm{latus}\mathrm{rectum}=\frac{39}{4} \end{gathered}$$