Question

Find the equation of the standard ellipse, taking its axes as the coordinate axes, whose minor axis is equal to the distance between the foci and whose length of latus rectum is $10$. Also, find its eccentricity.

Answer 1

Let $2a$ and $2b$ be the length of the axis of the ellipse.

Given, distance between foci$=$ length of minor axis

$2ae=2b$

$ae=b$

Also, $\frac{2b^2}{a}=10$

$b^2=5a$

$a^2{e}^2=5a$ [$\because$ $ae=b$]

$a=\frac{5}{e^2}$

Moreover $b^2=a^2(1-e^2)$

$a^2{e}^2=a^2(1-e^2)$

$e^2=1-e^2$

$2e^2=1$

$e^2=\frac{1}{2}$

From $a=\frac{5}{e^2}$

$a=5\times 2=10\Rightarrow$ $a^2=100$

$\therefore$ $b^2=5a=5\times 10=50$

$\therefore$ The equation of the ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\therefore \frac{x^2}{100}+\frac{y^2}{50}=1$

and $e=\frac{1}{\sqrt{2}}$