Question

Find the equation of an ellipse whose eccentricity is $\frac{2}{3}$, the latus-rectum is 5 and the centre is at the origin.

Answer 1

Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
The length of latus-rectum= 5
$\therefore \frac{2b^2}{a}=5$
$\Rightarrow b^2=\frac{5a}{2}$ ...(ii)
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow \frac{5a}{2}=a^2[1-{\left(\frac{2}{3}\right)}^2]$
$\frac{5a}{2}=a^2[1-\frac{4}{9}]$
$\Rightarrow \frac{5}{2}=a\left(\frac{5}{9}\right)$
$\Rightarrow \frac{5}{2}\times \frac{9}{5}=a$
$\Rightarrow a=\frac{9}{2}$
$a^2=\frac{81}{4}$
Putting $a=\frac{9}{5}$ in $b^2=\frac{5a}{2}$, we get
$b^2=\frac{5}{2}\times \frac{9}{2}$
$\Rightarrow b^2=\frac{45}{4}$
Substituting $a^2=\frac{81}{4}$ and $b^2=\frac{45}{4}$ in equation (i), we get
$\frac{x^2}{\frac{18}{4}}+\frac{y^2}{\frac{45}{4}}=1$
$\Rightarrow \frac{4x^2}{81}+\frac{4y^2}{45}=1$
This is the equation of the required ellipse.