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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.

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caylus
Hello,
If (a,b) is the focus and y=k is the directrice then the equation of the parabola is:
Something happended?
[tex]y= \dfrac{1}{2(b-k)} (x-a)^2+ \dfrac{b+k}{2} \\ Here\ a=-2, b=4,k=6\\ So\ y= \dfrac{-1}{4} (x+2)^2+5[/tex]


y=-x²/4-x+4
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

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