SOLUTION: Find the verticces for the hyperbola, -x^2+4y^2-6x-48y+139=0, and write your answer in this form: (x1,y1),(x2,y2).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the verticces for the hyperbola, -x^2+4y^2-6x-48y+139=0, and write your answer in this form: (x1,y1),(x2,y2).      Log On


   

Question 658628: Find the verticces for the hyperbola, -x^2+4y^2-6x-48y+139=0, and write your answer in this form: (x1,y1),(x2,y2).
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

-x^2+4y^2-6x-48y+139=0

x^2-4y^2+6x+48y =139

 (x+3)^2 - 4(y-6)^2 = 139 + 9 - 144

 (x+3)^2 - 4(y-6)^2 = 4

%28x%2B3%29%5E2%2F4+-+%28y-6%29%5E2%2F1+=+1

%28x%2B3%29%5E2%2F2%5E2+-+%28y-6%29%5E2%2F1%5E1+=+1 C(-3,6), V(-5,6) and (-1,6)

Standard Form of an Equation of an Hyperbola opening right and  left is:

  %28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1 with C(h,k) and vertices 'a' units right and left of center,