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Question 456433: 4x^2+10y^2-16x+40y+16=0 How do i find the center and foci?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 4x^2+10y^2-16x+40y+16=0 How do i find the center and foci?
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4x^2+10y^2-16x+40y+16=0
completing the square
4x^2-16x+10y^2+40y=-16
4(x^2-4x+4)+10(y^2+4+4)=-16+16+40
4(x-2)^2+10(y+2)^2=40
(x-2)^2/10+(y+2)^2/4=40
This equation is an ellipse with center at (2,-2), with a horizontal major axis.
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Standard form of ellipse with horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), with (h,k) being the (x,y) coordinates of the center.
Standard form of ellipse with vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), with (h,k) being the (x,y) coordinates of the center.
The difference between the two forms is the interchange of a^2 and b^2.
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For given equation:
a^2=10
b^2=4
c^2=a^2-b^2=10-4=6
c=√6
foci is at (2±√6,-2)
See the graph below as a visual check on the answer:
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y=(4-4(x-2)^2/10)^.5-2
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