Question 658628
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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
{{{(x+3)^2/4 - (y-6)^2/1 = 1}}}
{{{(x+3)^2/2^2 - (y-6)^2/1^1 = 1}}} C(-3,6), V(-5,6) and (-1,6)
Standard Form of an Equation of an Hyperbola opening right and  left is:
  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} with C(h,k) and vertices 'a' units right and left of center,