Question

Find the equation of the ellipse whose eccentricity is $\frac{3}{4}$, length of latus rectum is $6$, centre is $(0,0)$ and the major axis lies on $x-$axis

Answer 1

As the centre of the ellipse is $(0,0)$

i.e., origin and the major axis lies on $x-$axis, so its equation can be taken as

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

According to given,$\frac{2b^2}{a}=6\Rightarrow b^2=3a$

Also, eccentricity$=\frac{c}{a}=\frac{3}{4}\Rightarrow c=\frac{3a}{4}$

We know that $c^2=a^2-b^2$

$\Rightarrow {\left(\frac{3a}{4}\right)}^2=a^2-3a$

$\Rightarrow \frac{9a^2}{16}-a^2=-3a$

$\Rightarrow \frac{9a^2-16a^2}{16}=-3a$

$\Rightarrow \frac{-7a^2}{16}=-3a$

$\Rightarrow \frac{7a}{16}=3$

$\therefore a=\frac{48}{7}$

Put $a=\frac{48}{7}$ in $b^2=3a=3\times \frac{48}{7}=\frac{144}{7}$

From $\left(1\right)$ equation of the ellipse is

$\frac{x^2}{\frac{2304}{49}}+\frac{y^2}{\frac{144}{7}}=1$

$\Rightarrow \frac{49x^2}{2304}+\frac{7y^2}{144}=1$