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Question 81603: Write the equation of the ellipse in standard form. Find the center, vertices and foci: 4x^2 - 8x +9y^2 +36y + 8 = 0
I know I have to complete the square twice. Eventually I get to a point where I have 4(x - 4^2 + 9(y+2)^2 = 92. I then multiplied both sides by 1/92, so that I could get 1 on the left. But when I go to put it in standard form, I don't know what to do since the denominators can't be squared nicely, and I have a number stuck in the top of one of my numerators. I'm not sure If I'm doing this right. Can anyone please help me? Thanks.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Write the equation of the ellipse in standard form. Find the center, vertices and foci: 4x^2 - 8x +9y^2 +36y + 8 = 0
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4(x^2-2x+?) +9(y^2+4y+??) = -8
4(x^2-2x+1) + 9(y^2+4y+4) = -8 + 4 + 36
4(x-1)^2 + 9(y+2)^2 = 32
[(x-1)^2/8] + [(y+2)^2/(32/9)] = 1
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a=2sqrt2 ; b=[4sqrt(2)]/3
The c^2=a^2-b^2 = 8-(32/9)= 40/9
And c= [2sqrt(10)]/3
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Center at (1,-2)
Vertices at (1-2sqrt2,-2) and (1+2sqrt2)
Foci at (1-[2sqrt(10)]/3,-2) and (1+[2sqrt(10)]/3,-2)
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Cheers,
Stan H.
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