SOLUTION: -4x^2+9y^2-8x-54y+113=0, I need help in finding the CENTER,VERTICES,FOCI,and the equation that represents the ASYMPTOTES of the hyperbola

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: -4x^2+9y^2-8x-54y+113=0, I need help in finding the CENTER,VERTICES,FOCI,and the equation that represents the ASYMPTOTES of the hyperbola      Log On


   

Question 763701: -4x^2+9y^2-8x-54y+113=0, I need help in finding the CENTER,VERTICES,FOCI,and the equation that represents the ASYMPTOTES of the hyperbola
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
-4x%5E2%2B9y%5E2-8x-54y%2B113=0....write 113 as 81%2B36-4
-4x%5E2%2B9y%5E2-8x-54y%2B81%2B36-4=0......rearrange
4x%5E2%2B8x%2B4-+9y%5E2%2B54y-81%2B36=0..divide by -1
4x%5E2%2B8x%2B4-+9y%5E2%2B54y-81=36......group
%284x%5E2%2B8x%2B4%29-+%289y%5E2-54y%2B81%29=+36.......divide by 30
%284x%5E2%2F36%2B8x%2F36%2B4%2F36%29-+%289y%5E2%2F36-54y%2F36%2B81%2F36%29=+36%2F36...simplify
%28x%5E2%2F9%2B2x%2F9%2B1%2F9%29-+%28y%5E2%2F4-6y%2F4%2B9%2F4%29=+1
%281%2F9%29+%28x%5E2%2B2x%2B1%29-%281%2F4%29+%28y%5E2-6y%2B9%29=+1
%281%2F9%29+%28x%2B1%29%5E2-%281%2F4%29+%28y-3%29%5E2+=+1
%28x%2B1%29%5E2%2F9-%28y-3%29%5E2%2F4+=+1
the center is at (h, k)
h=-1 and k=3; so the center is at (-1, 3)

a=3, b=2, c%5E2+=+a%5E2%2Bb%5E2
c%5E2+=+9%2B4
c%5E2+=+13
c=sqrt%2813%29
vertices at (-4, 3) and (2, 3)
foci at (-1-sqrt%2813%29, 3) and (-1%2Bsqrt%2813%29, 3)
or (-4.6, 3) and (2.6, 3)
assymptotes at y=2x%2F3%2B11%2F3 and y=7%2F3-2x%2F3