What is the equation of a parabola with a focus at $(3, 2)$ $(3, 2)$ and a directrix at $y = - 4$ $y=-4$ ?

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Answer 1 · 2018-08-04T12:45:41

The equation of the parabola is $y = \frac{1}{12} {(x - 3)}^{2} - 1$ $y=1/12(x-3)^2-1$

The focus is $F = (3, 2)$ $F=(3,2)$ and the directrix is $y = - 4$ $y=-4$

Any point $(x, y)$ $(x,y)$ on the parabola is equidistant from the focus and the directrix.

$y + 4 = \sqrt{{(x - 3)}^{2} + {(y - 2)}^{2}}$ $y+4=sqrt((x-3)^2+(y-2)^2)$

Squaring both sides

${(y + 4)}^{2} = {(x - 3)}^{2} + {(y - 2)}^{2}$ $(y+4)^2=(x-3)^2+(y-2)^2$

$y^{2} + 8 y + 16 = {(x - 3)}^{2} + y^{2} - 4 y + 4$ $y^2+8y+16=(x-3)^2+y^2-4y+4$

$12 (y + 1) = {(x - 3)}^{2}$ $12(y+1)=(x-3)^2$

$y + 1 = \frac{1}{12} {(x - 3)}^{2}$ $y+1=1/12(x-3)^2$

The equation is $y = \frac{1}{12} {(x - 3)}^{2} - 1$ $y=1/12(x-3)^2-1$

graph{(y+1-1/12(x-3)^2)(y+4)=0 [-10, 10, -5, 5]}

What is the equation of a parabola with a focus at (3,2)(3, 2) and a directrix at y=−4y=-4?