Question

Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point (-3,1) and has eccentricity equal to $\sqrt{2/5}$.

Answer 1

Let the equation of the ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where a>b
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow b^2=a^2[1-{\left(\sqrt{\frac{2}{5}}\right)}^2]$
$\Rightarrow b^2=a^2[1-\frac{2}{5}]$
$\Rightarrow b^2=a^2\times \frac{3}{5}$
$\Rightarrow b^2=\frac{3a^2}{5}$ ...(i)
Ths required ellipse passes through (-3,1)
$\therefore \frac{(-3{)}^2}{a^2}+\frac{1^2}{b^2}=1$
$\Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1$ ...(ii)
Putting $b^2=\frac{3a^2}{5}$ in equation (ii), we get
$\frac{9}{a^2}+\frac{1}{\frac{3a^2}{5}}=1$
$\Rightarrow \frac{9}{a^2}+\frac{5}{3a^2}=1$
$\Rightarrow \frac{1}{a^2}[\frac{9}{1}+\frac{5}{3}]$
$\Rightarrow \frac{27+5}{3}=a^2$
$\to \frac{32}{3}=a^2$
$\Rightarrow a^2=\frac{32}{3}$
Putting $a^2=\frac{32}{3}$ in equation (ii), we get
$b^2=\frac{3}{5}\times \frac{32}{3}=\frac{32}{5}$
Substituting $a^2=\frac{32}{3}$ and $b^2=\frac{32}{5}$ in equation (i), we get,
$\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1$
$\Rightarrow \frac{3x^2}{32}+\frac{5y^2}{32}=1$
$\Rightarrow 3x^2+5y^2=32$
This is the equation of the required ellipse.