Question

Find the equation of the ellipse whose focii are $(4,0)$ and $(-4,0)$ and eccentricity is $\frac{1}{3}$

Answer 1

Since both focus lies on x-axis therefore x-axis is major axis and mid point of focii is origin which is centre and a line perpendicular to major axis and passes through centre is minor axis which is y-axisLet equation of ellipse is y-axis
Let equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\because ae=4ande=\frac{1}{3}\left(Given\right)$
$\therefore a=12and{b}^2=a^2(1-e^2)$
$\Rightarrow b^2=144(1-\frac{1}{9})\Rightarrow b^2=16\times 8\Rightarrow b=8\sqrt{2}$
Equation of ellipse is $\frac{x^2}{144}+\frac{y^2}{128}=1$