Find an equation for the ellipse that satisfies the given co

Find an equation for the ellipse that satisfies the given conditions:

Eccentricity 1/3, foci: (0, ±5)

Solution

The equation of ellipse is

x²/a²+y²/b²=1

Length of major axis,2a=26

a=13

And the ellipse passing through (sqrt13,6)

sqrt13²/13² +   6²/b²=1

36/b²=12/13

b²=39

Therefore the required equation is

x²/169 + y²/39 =1

Second question

The equation is of the form

y²/a² + x²/b² =1

Focii=(h,k+-c)

And in the given question (h,k)=(0,0)

Therefore c=5

And eccentricity ,e=c/a

1/3=5/a

a=15

c²=a²-b²

5²=15² -b²

b²=200

Therefore required equation is

y²/225 + x²/200=1