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Question 759821: Find the center and vertices of the ellipse x2 + 4y2 - 4x + 8y + 4 = 0
Answer by DSMLMD(16)   (Show Source):
You can put this solution on YOUR website! x^2 + 4y^2 - 4x + 8y + 4 = 0
x^2 - 4x + 4y^2 + 8y +4 =
(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = 0
(x-2)^2 - 4 + 4(y + 1)^2 = 0
(x - 2)^2 + 4(y + 1)^2 = 4
divided by 4 so we get the right side become 1 according to ellipse equation:
(x - 2)^2/4 + (y + 1)^2 = 1
The equation also can be written by:
(x - 2)^2/4 + (y + 1)^2/1 = 1
we get:
h=2
k=-1
a^2=4 -> a=2
b^2=1 -> b=1
center of ellipse:
(h,k)
(2,-1)
vertices of ellipse:
(aħh,k)
(2ħ2,-1)
(2+2,-1) and (2-2,-1)
(4,-1) and (0,-1)
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