SOLUTION: 9x^2 + 16y^2 - 18x + 64y - 71 = 0 find the coordinates of the center. a)(1,2) b)(1,-2) c)(-1,2) d)(-2,1)

Question 216624: 9x^2 + 16y^2 - 18x + 64y - 71 = 0
find the coordinates of the center.
a)(1,2)
b)(1,-2)
c)(-1,2)
d)(-2,1)
Found 2 solutions by RAY100, crystal_nguyen:
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
9x^2 -18x + 16y^2 +64y = 71
.
9(x^2 -2x ) + 16(y^2 +4y ) = 71,,,complete the squares remembering the last term = (2nd term / 2) squared,,,,,and add compensating amt to opposite side of eqn
.
9(x^2 -2x +1) +16(y^2 +4y+4) = 71 +(9*1) + (16*4) = 144
.
9(x-1)^2 +16(y+2)^2 =144,,,, divide by 144
.
(1/16){(x-1)^2} + (1/9){(y+2)^2} = 1
.
{(x-1)^2} / 16 + {(y+2)^2} / 9 =1,,,,,,,which is eqn for ellipse
.
Center is at (1,-2),,,,,,,which is answer (b)
.

Answer by crystal_nguyen(2) (Show Source): You can put this solution on YOUR website!
(9x^2-18x+9)+(16y^2+64y+64)+2=0
(3x-3)^2+(4y+8)^2=-2
No solution.
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