Question

Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point(-3,1) and has eccentricity $\sqrt{\frac{2}{5}}$.

Answer 1

Let the equation of the required ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{\frac{2}{5}}=\sqrt{1-\frac{a^2}{b^2}}$
$\Rightarrow \sqrt{\frac{2}{5}}=\sqrt{1-\frac{a^2}{b^2}}$
$\Rightarrow \frac{2}{5}=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{b^2}{a^2}=1-\frac{2}{5}$
$\Rightarrow \frac{b^2}{a^2}=\frac{3}{5}$
$\Rightarrow 5b^2=3a^2$
$\Rightarrow b^2=\frac{3a^2}{5}$ ...(ii)
Putting the value of $b^2=\frac{3a^2}{5}$ in equation (ii),
we get
$\frac{9}{a^2}+\frac{1}{\frac{3a^2}{5}}=1$
$\Rightarrow \frac{9}{a^2}+\frac{5}{3a^2}=1$
$\Rightarrow \frac{1}{a^2}[9+\frac{5}{3}]=1$
$\Rightarrow 9+\frac{5}{3}=a^2$
$\Rightarrow a^2=\frac{32}{3}$ ...(iii)
Putting $a^2=\frac{32}{3}$ in equation (ii), we get
$b^2=\frac{3}{5}\times \frac{32}{3}=\frac{32}{5}$ ...(iv)
$\therefore$ The required equation of ellipse is
$\frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1$
$\Rightarrow \frac{3x^2}{32}+\frac{5y^2}{32}=1$
$\Rightarrow 3x^2+5y^2=32$
This is the required equation of ellipse.