Question

Find the equation of an ellipse whose latus rectum is $8$ and eccentricity is $\frac{1}{3}$.

Answer 1

For the ellipse $\frac{x^2}{y^2}+\frac{y^2}{b^2}=1$

Latus rectum $\left(\begin{array}{c}\frac{2b^2}{a}\end{array}\right)=8$ and eccentricity $\left(e\right)=\frac{1}{3}$

$\therefore \frac{2b^2}{a}=8$ ...(i) [where $b^2=a^2(1-e^2)$]

$\Rightarrow \frac{2a^2(1-e^2)}{a}=8$

$\Rightarrow 2a\left\{\begin{array}{c}1-{\left(\begin{array}{c}\frac{1}{3}\end{array}\right)}^2\end{array}\right\}=8$

$\Rightarrow a\left(\begin{array}{c}1-\frac{1}{9}\end{array}\right)=4$

$\Rightarrow a\left(\begin{array}{c}\frac{8}{9}\end{array}\right)=4$

$\therefore a=\frac{9\times 4}{8}=\frac{9}{2}$

For equation (i), $2b^2=8a$

$b^2=4a=4\times \frac{9}{2}=18$

Hence, the required equation of the ellipse is

$\frac{x^2}{{\left(\frac{9}{2}\right)}^2}+\frac{y^2}{18}=1$

$\Rightarrow \frac{4x^2}{81}+\frac{y^2}{18}=1$.