SOLUTION: ``Find the vertices,asymptotes, and the foci of the hyperbola. y^(2)-9x^(2)-12y-36x-9=0.

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Question 633559: ``Find the vertices,asymptotes, and the foci of the hyperbola.
y^(2)-9x^(2)-12y-36x-9=0.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

y^(2)-9x^(2)-12y-36x-9=0   ⇒ (y-6)^2 - 9(x+2)^2 = 9

%28y-6%29%5E2%2F3%5E2+-+%28x%2B2%29%5E2%2F1%5E2+=+1 C(-2,6)

 V(-2, 9) and V(-2,3)

F%5Bd%5D+=+sqrt%2810%29  Foci (-2, 6 ±sqrt%2810%29) 

Asymptotes: 

     y-6 = 3(x +2)   y = 3x+12

and  y-6 = -3(x+2)   y = -3x 

Standard Form of an Equation of an Hyperbola opening up and down is:

  %28y-k%29%5E2%2Fb%5E2+-+%28x-h%29%5E2%2Fa%5E2+=+1 with C(h,k) and vertices 'b' units up and down from center,  2b the length of the transverse axis

Foci sqrt%28a%5E2%2Bb%5E2%29units units up and down from center, along x = h

& Asymptotes Lines passing thru C(h,k), with slopes m =  ± b/a