How do I find the center of an ellipse with the equation $9x^2+16y^2-18x+64y=71$ ?

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Answer 1 · 2014-10-07T17:17:33

Group the variables first

$9X^2 - 18X + 16Y^2 + 64Y = 71$

Do some factoring

$9(X^2 - 2X) + 16(Y^2 + 4Y) = 71$

Now, add something X and Y perfect squares.

$9(X^2 - 2X + 1) + 16(Y^2 + 4Y + 4) = 71$

However, since we added something on the left side of the equation,
we need to add the same value on the right side to maintain the equation.

$9(X^2 - 2X + 1) + 16(Y^2 + 4Y + 4) = 71 + 9(1) + 16(4)$

$9(X - 1)^2 + 16(Y + 2)^2 = 144$

From here it's already obvious that the ellipse is centered at (1, -2)

How do I find the center of an ellipse with the equation 9x^2+16y^2-18x+64y=719x^2+16y^2-18x+64y=71?