Question 421164: 4x^2+25y^2-8x+100y+4=0 find the center, vertices and foci of the ellipse
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 4x^2+25y^2-8x+100y+4=0 find the center, vertices and foci of the ellipse
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Standard form of the given ellipse: (x-h)^2/a^2+(y-k)^2/b^2=1 (a>b)
4x^2+25y^2-8x+100y+4=0
4x^2-8x+25y^2+100y+4=0
factor then complete the square
4(x^2-2x+1)+25(y^2+4y+4)=-4+4+100
4(x-1)^2+25(y+2)^2=100
divide by 100
(x-1)^2/25+(y+2)^2/4=1
This is an ellipse with center at(1,-2) and major axis horizontal
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a^2=25
a=5
b^2=4
b=2
c^2=a^2-b^2
c=sqrt(25-16)=sqrt(9)=3
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foci:(1+-c,-2)=(1+-3,-2)=(4,-2)and (-2,-2)
vertices:(1+-a,-2)=1+-5,-2)=(6,-2) and (-4,-2)
See graph of the ellipse below:
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y=-2+-(1-(x-1)^2/25)^.5
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