find the equation of a parabola whose focus is at (0,-2) and directrix at y=4

Question

I need help finding the equation for the parabola

Answer 1

You can use definition that from any point (x, y) on the curve to the focus is equal distant to the directrix.

x^2 + (y+2)^2 = (y-4)^2

=>x^2 + y^2 + 4y + 4 = y^2 - 8y + 16

Answer: y = (-1/12)(x^2) + 1

 

Or, you can use vertex form:

The vertex is at (0,  (-2+4)/2) = (0, 1)

2a = 4+2 = 6, where 2a is the distance between the focus and the dirextrix.

Since it opens down, we have

y = (-1/(4a))x^2 + 1 = (-1/12)(x^2) + 1

Answer 2

The vertex is always halfway between the focus and the directrix, and the parabola always curves away from the directrix.

 

so if the focus is at (0,-2) and the directrix is at y = 4

the vertex would be halfway between here

 

So the vertex, exactly between the focus and directrix, must be at (h, k) = (0, 1). The absolute value of p is the distance between the vertex and the focus and the distance between the vertex and the directrix. (The sign on p tells me which way the parabola faces.) Since the focus and directrix are six units apart, then this distance has to be three units, so | p | = 3.

 

now to determine the direction the parbola is going to curve it is going to curve away from the directix which is a line at 4 parallel to the x axis

 

since the vertex is at (0,1) the parabola will curve down which tells me this is a regular parabola so the eqn is

 

(x – h)² = 4p(y – k)

 

we know h,k and p (p is = -|p| because it is pointing down

 

(x-0)² = 4(-3)(y-1)

 

x² = -12(y-1)

 

now if your directix was parallel to the y axis you would have used the formula

 

(x – h)2 = 4p(y – k)

 

see http://www.purplemath.com/modules/parabola.htm for more help