What is the equation in standard form of the parabola with a focus at (12,-3) and a directrix of y= 12?

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Answer 1 · 2017-09-03T10:09:44

$y-9/2=-1/30(x-12)^2.$

We will use the following Focus-Directrix Property of Parabola (FDP) :

FDP : Any point on a Parabola is equidistant from the

Focus F and the Directrix d.

Let $P(x,y)$ be any arbitrary point on the Parabola. Then.

The Distance $PF,$ btwn. #P(x,y), and, F(12,-3), is,

$PF=...............................(1).$

The Dist. btwn $P(x,y), and, d : y-12=0," is, "|y-12|...(2).$

$:. (1), (2),$ and, FDP,

$rArr sqrt{(x-12)^2+(y-(-3)^2)}=|y-12|.$

$:. (x-12)^2=|y-12|^2-(y+3)^2.$

$:. (x-12)^2=y^2-24y+144-y^2-6y-9, i.e.,$

$(x-12)^2=-30y+135=-30(y-9/2).$

$:. y-9/2=-1/30(x-12)^2,$ is the desired eqn. of the Parabola.