Question

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

Answer 1

Critical points:
`color(white)("XXX")(-6,0), (+6,0), (0,-4), (0,+4)`

Explanation:

The general form for an ellipse is
`color(white)("XXX")((x-a)^2)/h^2 + ((y-b)^2)/k^2 =1`
with
`color(white)("XXX")`center at `(a,b)`,
`color(white)("XXX")`x-axis radius `h`, and
`color(white)("XXX")`y-axis radius `k`

Converting the given equation to this form, we have
`color(white)("XXX")((x-0)^2)/(6^2) +((y-0)^2)/(4^2)=1`

Giving us critical points on the Y-axis of `x=+-6`
and critical points on the X-axis of `y=+-4`
graph{x^2/36+y^2/16=1 [-8.89, 8.89, -4.444, 4.44]}