Parabola in standard form

Question

Derive the equation of the parabola with a focus at (4,-7) and a director of y = -15. Put the equation in standard form

Answer 1

focus (4,-7), directrix y=-15

the vertex is halfway between the focus and directrix, so its y coordinate is (-7-15)/2 = -11

the vertex is (x,-11) -11 is a distance of 4 to both the focus and directrix

the parabola is upward opening with a>0, for the general equation y=a(x-h)^2 + k where (h,k) is the vertex, with k=-11, h=4

y=a(x-4)^2 -11

plug in another point on the parabola

another point will also be equidistant to the focus and directrix, such as (12,-7) directly to the right of the focus. It's a distance of 8 to the directrix, so it's also 8 units to the right of the focus, with x coordinate 4+8 = 12

-7 = a(12-4)^2 +11

solve for a

a=1/16

the parabola's equation is: y= (1/16)(x-4)^2 - 11