Question

Equation of the ellipse with foci $(\pm 4,0)$ and length of latus rectum $\frac{20}{3}$ is

• A. $\frac{x^2}{20}+\frac{y^2}{36}=1$
• B. $\frac{x^2}{20}+\frac{y^2}{4}=1$
• C. $\frac{x^2}{36}+\frac{y^2}{20}=1$
• D. $\frac{x^2}{24}+\frac{y^2}{8}=1$

Answer 1

The correct option is C $\frac{x^2}{36}+\frac{y^2}{20}=1$

We know foci of the ellipse is ($\pm ae$,0)

So, $ae=4$

and, length of latus rectum = $\frac{{2b}^2}{a}$

So, $\frac{{2b}^2}{a}$ = $\frac{20}{3}$----------------(1)

and we know $b^2=a^2-{\left(ae\right)}^2$

or, $b^2=a^2-16$

$\therefore \frac{2(a^2-16)}{a}=\frac{20}{3}$

or, $3a^2-10a-48=0$

On solving the quadratic we get two values one positive and one negative.

Neglecting the negative value,

$\therefore a=6$

Putting the value of a=6 in equation (1)we get,

$\frac{2b^2}{6}=\frac{20}{3}$

$\therefore b^2=20$

$\therefore$ Equation of ellipse is : $\frac{x^2}{36}+\frac{y^2}{20}=1$