Consider the ellipse 4x^2+49y^2=36
I can figure the vertices is (3,0)(-3,0)
I cannot figure out the foci
Thanks
4x^2/36 + 49y^2/36 = 1
x^2/9 + y^2/(36/49) = 1
a = 3
b = 6/7
c^2 = a^2-b^2 = 9-36/49 = 405/49
c = 9√5/7
You found the vertices ok.
They are at (±a,0)
The foci are at (±c,0)