Need ! find the equation of the hyperbola centered at the origin that satisfies the given conditions: vertex: (0, square root of 12), passing through (2, square root of 3, 6) a. y^2/36 - x^2/12 =1 b. y^2/12 -x^2/36 =1 c. y^2/6 - x^2/12 =1 d. y^2/12 - x^2/6 =1

You know from the y-intercept that the equation will have a +y^2/12 term in it. Then for y=6, you have
.. 6^2/12 -x^2/b = 1
.. 2 = x^2/b
.. b = x^2/2
If your point is (2√3, 6), then this is
.. b = (2√3)^2/2 = 12/2 = 6

Then the hyperbola's equation is
.. y^2/12 -x^2/6 = 1 . . . . . . . . selection D

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4