A parabola can be drawn given a focus of (5,4) and a directrix of x=7. What can be said about the parabola

Answer 1

It opens to the left, with vertex (6,4)

it's general form is x-h = a(y-k) where (h,k) is the vertex = (6,4) (6 is the average of 5 and 7)

x-6 = a(y-4)^2

find another point on the parabola, plug it in to solve for a, the coefficient of the y^2 term

find a point directly above the focus (5,h) h is equidistant to the focus and directrix, h-4 = 7-5=2, h=6

(5,6) is on the parabola

5-6 = a(6-4)^2

-1 = 4a

a = -1/4

the parabola is

x-6= -(1/4)(y-4)^2 or

4x = -y^2 + 8y -10