Question

Find the equation for the ellpse that satisfies the given conditions
Vertices $(0,\pm 13),Foci(0,\pm 5)$

Answer 1

The foci $(0,\pm 5)$ lie on y- axis.
So the equation of ellipse in standard form is
$\frac{y^2}{a^2}+\frac{x^2}{b^2}=1\text{now vertices}(0,\pm a)\text{is}(0,\pm 13)\Rightarrow a=13\text{foci}(0,\pm c)\text{is}(0,\pm 5)\Rightarrow c=5\text{We know that}{c}^2=a^2-b^2\therefore (5{)}^2=(13{)}^2-b^2\Rightarrow b^2=169-25=14\text{Thus equation of required ellipse is}\frac{x^2}{144}+\frac{y^2}{169}=1.$