Question

Find the equation of the ellipse in the standard form whose minor axis is equal to the distance between foci and whose latus-rectum is 10.

Answer 1

The coordinates of foci are $(\pm ae,0)$
$\therefore$ 2ae=2b [given]
$\Rightarrow ae=b$
$\Rightarrow (ae{)}^2=b^2$ ...(i)
The length of latus-rectum is 10.
$\Rightarrow \frac{2b^2}{a}=10$
$\Rightarrow b^2=\frac{10a}{2}$
$\Rightarrow b^2=5a$ ...(ii)
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow b^2=a^2-a^2{e}^2$
$\Rightarrow b^2=a^2-b^2$
$\Rightarrow 2b^2=a^2$
$\Rightarrow b^2=\frac{a^2}{2}$
Substituting $b^2=\frac{a^2}{2}$ in equation (ii), we get
$\frac{a^2}{2}=5a$
$\Rightarrow a^2=10a$
$\Rightarrow a=10$
$\Rightarrow a^2=100$
Putting $a^2=100$ in $b^2=\frac{a^2}{2},$ we get
$b^2=\frac{100}{2}=50$
$\therefore$ The required equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{100}+\frac{y^2}{50}=1$
$\Rightarrow \frac{x^2+2y^2}{100}=1$
$\Rightarrow x^2+2y^2=100$
This is the required equation of the ellipse.