How to find the co-vertices and vertices of (x^2)/9 + (y^2)/36 ?

Answer 1

if that's supposed to be x²/9 + y²/36 = 1, the equation of an ellipse, and if by vertices and co-vertices you mean x and y intercepts, then you get those by letting y and x be zero and then solving:

x²/9 + 0 = 1
x² = 9
x = ±3

similarly, y = ±6

foci are on the longer axis at √(36-9) = √27 = 3√3 ≈ ±5.19