Hi I have a 2 part question 1st I need to write (y-2)^2 -x^2/4 =1 in general form then find the centre and vertices of the hyperbola. The answer I got was -x^2 +4y^2-16y+12=0 is this right.

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Your general form is correct. For the center, think of the x^2 as (x - 0)^2 so the center is (0, 2) not (1, 2) To find the vertices you need to know that the vertices are "a" units from the center. "a^2" is always under the positive term. In this problem, "a^2" is the denominator of the (y - 2)^2 term. Since there is no denominator, it is understood to be 1. Add and subtract 1 to the y coordinate of the center to get (0, 1) and (0, 3) for the vertices.