Question 1032832
From
{{{13x^2+3y^2=39}}}
we divide by 39 and get
{{{x^2/3 + y^2/13 = 1}}}
This is a "tall" ellipse.
c^2 = 13 - 3 = 10
so that the foci are at
(0, sqrt(10)) and (0, -sqrt(10))