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

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Answer 1 · 2016-12-14T07:49:36

The center is #=(0,0)#
The vertices are #(1/2,0)#,#(-1/2,0)#,#(0,1)# and #(0,-1)#

Let's rewrite the equation

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

The equation of the ellipse, center, #(h,k)#

is

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

So, the center is #(0,0)#

To find the vertices, let's #y=0#

Then, #x=+-1/2#

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

Let #x=0#, #=>#, #y=+-1#

The other vertices are #(0,1)# and #(0,-1)#
graph{4x^2+y^2=1 [-2.16, 2.165, -1.082, 1.08]}