What are the critical points of #x^2/9 + y^2/36 = 1#?

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Answer 1 · 2016-11-10T07:29:32

The axes are #=12 and 6#
The center is #(0,0)#
The foci are F#=(0,5)# and #(0,-5)#

This is the equation of an ellipse.
The major axis #=12#
The minor axis #=6#

#(x-0)^2/3^2+(y-0)^2/6^2=1#
The center is #(0,0)#

#c=sqrt(36-9)=sqrt25=5#

The foci are F#=(0,5)# and #(0,-5)#

graph{(x^2/9)+(y^2/36)=1 [-14.24, 14.23, -7.12, 7.12]}

Answer 2 · 2016-11-10T07:30:18

Critical points are #(0,6)#, #(0,-6)#, #(3,0)# and #(-3,0)#.

A critical point is a point on a curve, where either the derivative is not defined or its value is zero.

As #x^2/9+y^2/36=1#, we have

#(2x)/9+(2y)/36xx(dy)/(dx)=0#

or #(dy)/(dx)=-(2x)/9xx36/(2y)=-(4x)/y#,

which is zero when #x=0# (#y#-axis) and is not defined when #y=0# (#x#-axis).

Note that when #x=0#, #y=+-6# and when #y=0#, #x+-3#, i.e.

Critical points are #(0,6)#, #(0,-6)#, #(3,0)# and #(-3,0)#.

In fact, it is an ellipse with major axis at #y#-axis and minor axis at #x#-axis.
graph{x^2/9+y^2/36=1 [-20, 20, -10, 10]}