Question

Find the equation of the ellipse whose centre lies at the origin, major axis lies on the x-axis, the eccentricity is $\frac{2}{3}$ and the length of the latus rectum is 5 units.

Answer 1

Since the major axis of the ellipse lies on the x-axis, it is a horizontal ellipse.

Let the required equation be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2>b^2.$

Length of its latus rectum = $\frac{2b^2}{a}.$

$\therefore \frac{2b^2}{a}=5\iff \frac{2a^2(1-e^2)}{a}[\because b^2=a^2(1-e^2\left)\right]$

$\iff a=\frac{5}{2(1-e^2)}=\frac{5}{2(1-\frac{4}{9})}=(\frac{5}{2}\times \frac{9}{5})=\frac{9}{2}.$

Also, $b^2=a^2(1-e^2)=\frac{81}{4}\times (1-\frac{4}{9})=(\frac{81}{4}\times \frac{5}{9})=\frac{45}{4}.$

$\therefore a^2={\left(\frac{9}{2}\right)}^2=\frac{81}{4}$ and $b^2=\frac{45}{4}.$

Hence, the required equation is

$\frac{x^2}{81/4}+\frac{y^2}{45/4}\iff \frac{4x^2}{81}+\frac{4y^2}{45}=1\iff 20x^2+36y^2=405.$