Question

Help with Parabolas? :/
Find the standard form of the equation of the parabola with a focus at (0, -9) and a directrix y = 9.

Answer 1

@whpalmer4

Answer 2

@phi

Answer 3

start by drawing a diagram of the information



so looking at the diagram, because the dirextrix is above the focus, the parabola is concave down...

so you need to find the vertical distance from the focus to the dirextrix...

Answer 4

So 18? @campbell_st

Answer 5

yep...

so the distance you just found is twice the focal length... and denoted by 2a

where a is the focal length..

the vertex of the parabola is located a units above the focus

so the vertex is at (0, -9 + 9) = (0,0)

so now you know the vertex, focal length and concavity

I use

then I use this form of the parabola

\[(x - h)^2 = 4a(y - k)\]

where the vertex is (h, k) and the focal length is a

since the vertex is above the focus I substitute x = -a

so the equation is

\[(x - 0)^2 = 4\times(-9)(y -0)\]

or

x^2 = -36y

then

in standard form

\[y = -\frac{x^2}{36}\]

I just find this easier than using the distance formula between a point on the parabola(x, y) and the focus then equating it to the distance from the point (x, y) to the direactrix.

Answer 6

if you want the 2nd method its



by definition the distance AP = FP...

so use the distance formula... to get the equation