Question

Find the eccentricity, coordinates of foci, length of the latus-rectum of the ellipse: $4x^2+9y^2=1$

Answer 1

Equation of ellipse: $4x^2+9y^2=1$

$\Rightarrow \frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{9}}=1$

$\therefore a^2=\frac{1}{4}$ and $b^2=\frac{1}{9}$

For eccentricity,$\left(e\right)=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{\frac{1}{9}}{\frac{1}{4}}}$

$\Rightarrow e=\sqrt{1-\frac{4}{9}}$

$\Rightarrow e=\frac{\sqrt{5}}{3}$

For coordinates of foci

$\because a^2=\frac{1}{4}\Rightarrow a=\frac{1}{2}$

$ae=\frac{1}{2}\times \frac{\sqrt{5}}{3}=\frac{\sqrt{5}}{6}$

So, the foci are $(\pm ae,0)=(\pm \frac{\sqrt{5}}{6},0)$

Length of latus−rectum $=\frac{2b^2}{a}=\frac{2\times \frac{1}{9}}{\frac{1}{2}}=\frac{4}{9}$