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

1 Answer
Dec 13, 2015

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]}