Question

Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form.Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form.

Answer 1

for the first parabola the axis fs symmetry is parallel to y axis and the y coordinate of the vertex is halfway between y = 6 and y = 4. The paraloa opens downward.

the general formula is ( x - h)^2 = -4p(y - k) where (h,k is the vertex and p = distance of the focuse from the vertex which in this case is 1.
s we have

(x + 2)^2 = -4 ( y- 5)

x^2 + 4x + 4 = -4y + 20

y = -0.25x^2 - x + 4 is the required equation

Answer 2

Hello,
If (a,b) is the focus and y=k is the directrice then the equation of the parabola is:
Something happended?

`y= (1)/(2(b-k)) (x-a)^2+ (b+k)/(2) \\ Here\ a=-2, b=4,k=6\\ So\ y= (-1)/(4) (x+2)^2+5`


y=-x²/4-x+4