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

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2 Answers

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
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)