# How do you find the critical points for #(x^2)/36 + (y^2)/16 =1#?

##### 1 Answer

Dec 13, 2015

Critical points:

#### Explanation:

The general form for an ellipse is

with

Converting the given equation to this form, we have

Giving us critical points on the Y-axis of

and critical points on the X-axis of

graph{x^2/36+y^2/16=1 [-8.89, 8.89, -4.444, 4.44]}