Question 422333
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Hi
Finding the center of this hyperbola: 
 -4x^2+9y^2+16x-18y+29=0
 9(y^2-2y)  - 4(x^2-4x) + 29 = 0  |Completing the square
9[(y-1)^2-1  - 4[(x-2)^2 -4] + 29 = 0
9(y-1)^2 -9 - 4(x-2)^2 +16 + 29 = 0
9(y-1)^2 - 4(x-2)^2 +36 = 0
 4(x-2)^2 - 9(y-1)^2 = 36
 {{{(x-2)^2/9 - (y-1)^2/4 = 1}}}  Center Pt(2,1)
{{{drawing(300,300,   -6, 6, -6, 6, grid(1),
circle(2, 1,0.3),
graph( 300, 300, -6, 6, -6, 6,0, 2sqrt(((x-2)^2/9) -1)+1,-2sqrt(((x-2)^2/9) -1)+1))}}}