The directrix of the parabola 12(y+3)=(x-4)^2…

Question

The directrix of the parabola 12(y+3)=(x-4)^2 has the equation y=-6. Find the coordinates of the focus of the parabola. This is question 30 on the June 2016 algebra 2 cc regents.

Answer 1

There is actually more information given in this problem that is needed, because the directrix equation isn't necessary; you can determine p from the coefficient, 4p, of the linear side: 12p(y+3) = (x-4)².

 

therefore 12 = 4p; p = 3.  You have its vertex at (h,k) = (4,-3).

 

You must move directly upward from vertex, p=3 units, to arrive at focus: F(4,0)

& just move directly dowward from vertex, p=3 units, to arrive on the directrix line, y = -3-3 = -6.