# What are the foci of the ellipse x^2/49+y^2/64=1?

Mar 2, 2015

The answer is: ${F}_{1 , 2} \left(0 , \pm \sqrt{15}\right)$.

The standard equation of an ellipse is:

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$.

This ellipse is with the foci (${F}_{1 , 2}$) on the y-axis since $a < b$.

So the ${x}_{{F}_{1 , 2}} = 0$

The ordinates are:

$c = \pm \sqrt{{b}^{2} - {a}^{2}} = \pm \sqrt{64 - 49} = \pm \sqrt{15}$.

So:

${F}_{1 , 2} \left(0 , \pm \sqrt{15}\right)$.