Find Parabola Features given Graph

Question

A parabola can be drawn given a focus of ( 0 , − 2 ) and a directrix of x = − 6 . What can be said about the parabola?

Answer 1

it's a rightward opening parabola with vertex having an x coordinate of -3 = the average of 0 and -6

y coordinate of the vertex = -2

the parabola's equation in vertex form is'

x= a(y+2)^2 -3

find one other point on the parabola, plug it into that equation to solve for a, the coefficient of the y^2 term

it may help to draw a rough sketch of the parabola, the directrix, a vertical line x=-6, the focus (0,-2) and vertex (-3,-2)

(0, 4) is a point on the parabola, directly above the focus

6 units above the focus and 6 units right from the directrix

plug x=0, y=4 to solve for "a"

x =a(y +2)^2 -3

0 =a(4+2)^2 -3

36a = 3

a = 3/36 = 1/13

the parabola is

x = (1/13)(y+2)^2 -3

or

13x= y^2 +4y + 1