Which is the equation of a hyperbola centered at the origin with vertex (0, sqrt 12 ) that passes through (2 sqrt 3,6)

y-intercept = 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

Step-by-step explanation:

answer: x=60

oof

step-by-step explanation: