How do you find the center and vertices of the ellipse #x^2/4+y^2/(1/4)=1#?

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Answer 1 · 2016-12-30T06:29:57

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

The vertices are #B=(0,1/2)# and #B'=(0,-1/2)#

We compare this equation to the general equation of an ellipse

#(x-h)^2/a^2+(y-k)^2/b^2=1#

#x^2/2^2+y^2/(1/4)=1#

The center of the ellipse is #h,k)=(0,0)#

To find the vertices

Let #y=0#, #=>#, #x^2=4#, #=>#, #x=+-2#

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

Let #x=0#, #=>#, #y^2=1/4#, #=>#, #y=+-1/2#

The vertices are #B=(0,1/2)# and #B'=(0,-1/2)#

graph{(x^2/4+y^2/(1/4)-1)=0 [-2.43, 2.435, -1.216, 1.217]}