SOLUTION: Find the vertices,asymptotes, and the foci of the hyperbola. y^(2)-4x^(2)-6y-32x-59=0

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Question 633543: Find the vertices,asymptotes, and the foci of the hyperbola.
y^(2)-4x^(2)-6y-32x-59=0
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Re TY, Yes!  Good Work!

y^(2)-4x^(2)-6y-32x-59=00   ⇒ (y-3)^2 - 4(x+4)^2 = 4

%28y-3%29%5E2%2F2%5E2+-+%28x%2B4%29%5E2%2F1%5E2+=+1 C(-4,3)

 V(-4, 5) and V(-4,1)

F%5Bd%5D+=+sqrt%285%29  Foci (-4, 3 ±sqrt%285%29) 

Asymptotes: 

     y-3 = 2(x +4)   y = 2x +11  

and  y-3 = -2(x+4)   y = -2x -5

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