You can put this solution on YOUR website! find the center of this ellipse:
16x^2+4y^2+64x+24y+36 = 0
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16(x^2+4x+4) + 4(y^2+6y+9) = -36+16*4+4*9
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16(x+2)^2 + 4(y+3) = 24
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Center at (-2,-3)
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Cheers,
Stan H.
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Hi
Standard Form of an Equation of an Ellipse is
where Pt(h,k) is the center and a and b are the respective vertices, distances from center.
Finding the center of the following ellipse:
16x^2+4y^2+64x+24y+36=0
16(x^2 + 4x) + 4(y^2+6y)+36 = 0 |completing the squares
16[(x+2)^2 -4] + 4[(y+3)^2-9] + 36 = 0
16(x+2)^2 - 64 + 4(y+3)^2 - 36 + 36 = 0
16(x+2)^2 + 4(y+3)^2 = 64
center pt(-2,-3)
(