Question

Find the equation of ellipse whose eccentricity is $\frac{2}{3}$, latus rectum is $5$ and the centre is $(0,0)$.

Answer 1

Let the equation of ellipse be of the form:

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

Given, latus rectum of ellipse is $5$

$\because$ lactus rectum$=\frac{2b^2}{a}$

$\Rightarrow \frac{2b^2}{a}=5$

$\Rightarrow b^2=\frac{5a}{2}$ …(2)

We know that, $b^2=a^2(1-e^2)$

$\Rightarrow \frac{5a}{2}=a^2(1-\frac{4}{9})$

$[\because$ Given, $e=\frac{2}{3}$ ]

$\Rightarrow \frac{5}{2}=a\left(\frac{5}{9}\right)$

$\Rightarrow a=\frac{9}{2}$

$\Rightarrow a^2=\frac{81}{4}$

$\Rightarrow b^2=\frac{5}{2}\times \frac{9}{2}=\frac{45}{4}$

[Using equation (2)]

Put the value of $a^2$ and $b^2$ in equation (1),

So, required equation of ellipse is:

$\frac{4x^2}{81}+\frac{4y^2}{45}=1$