Question

The equation of the ellipse with vertices $(\pm 13,0)$ and foci $(\pm 5,0),$ is

• A. $\frac{x^2}{169}+\frac{y^2}{144}=1$
• B. $\frac{x^2}{13}+\frac{y^2}{12}=1$
• C. $\frac{x^2}{13}+\frac{y^2}{11}=1$
• D. $\frac{x^2}{169}+\frac{y^2}{121}=1$

Answer 1

The correct option is A $\frac{x^2}{169}+\frac{y^2}{144}=1$

Given,
Vertices $\equiv (\pm 13,0)=(\pm a,0),$
Foci $\equiv (\pm 5,0)=(\pm ae,0)$

The equation of the required ellipse will be of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$
$\therefore a=13$
and $ae=5$
$\Rightarrow a^2{e}^2=25(\because e=\sqrt{1-\frac{b^2}{a^2}})\Rightarrow a^2-b^2=25\Rightarrow b^2=144$

Hence the equation of the ellipse is $\frac{x^2}{169}+\frac{y^2}{144}=1.$