Question

Find the equation of the ellipse whose foci are (4,0) and (-4,0), eccentricity = 1/3.

Answer 1

Let the equation of the required ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
The coordinates of foci are (+ae,0) and (-ae,0)
$\therefore$ ae =4
$\Rightarrow a\times \frac{1}{3}=4$
$\Rightarrow a=12\Rightarrow a^2=144$
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow b^2=144[1-{\left(\frac{1}{3}\right)}^2]$
$\Rightarrow b^2=144[1-\frac{1}{9}]$
$\Rightarrow b^2=144\times \frac{8}{9}$
$\Rightarrow b^2=16\times 8=128$
Substituting $a^2=144$ and $b^2=128$ in equation (i), we get
$=\frac{x^2}{144}+\frac{y^2}{128}=1$
$\Rightarrow \frac{1}{16}[\frac{x^2}{9}+\frac{y^2}{8}]=1$
$\Rightarrow \frac{x^2}{9}+\frac{y^2}{8}=16$
This is the equation of the required ellipse.