How do you find the center and vertices of the ellipse #4x^2+9y^2=36#?

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Answer 1 · 2017-08-03T05:05:58

The center is #=(0,0)#.
The vertices are #A'=(-3,0)#,#A= (3,0)#, #B'=(-2,0)# and #B=(2,0)#

The equation is

#4x^2+9y^2=36#

Divide throughout by #36#

#4/36x^2+9/36y^2=36/36#

#x^2/9+y^2/4=1#

#x^2/3^2+y^2/2^2=1#

This is the standard equation of an ellipse center #=(0,0)#

When #y=0#

#x^2/3^2=1#, #=>#, #x^2=9#, #=>#, #x=+-3#

The vertices are #A'=(-3,0)# and #A=(3,0)#

When #x=0#

#y^2/2^2=1#, #=>#, #y^2=4#, #=>#, #x=+-2#

The vertices are #B'=(-2,0)# and #B=(2,0)#

graph{(x^2/9)+(y^2/4)=1 [-8.89, 8.89, -4.444, 4.445]}