What is the equation in standard form of the parabola with a focus at (14,5) and a directrix of y= -3?

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Answer 1 · 2017-08-08T13:39:05

The equation of the parabola is ${(x - 14)}^{2} = 16 (y - 1)$ $(x-14)^2=16(y-1)$

Any point $(x, y)$ $(x,y)$ on the parabola is equidistant from the focus $F = (14, 5)$ $F=(14,5)$ and the directrix $y = - 3$ $y=-3$

Therefore,

$\sqrt{{(x - 14)}^{2} + {(y - 5)}^{2}} = y + 3$ $sqrt((x-14)^2+(y-5)^2)=y+3$

${(x - 14)}^{2} + {(y - 5)}^{2} = {(y + 3)}^{2}$ $(x-14)^2+(y-5)^2=(y+3)^2$

${(x - 14)}^{2} + y^{2} - 10 y + 25 = y^{2} + 6 y + 9$ $(x-14)^2+y^2-10y+25=y^2+6y+9$

${(x - 14)}^{2} = 16 y - 16 = 16 (y - 1)$ $(x-14)^2=16y-16=16(y-1)$

graph{((x-14)^2-16(y-1))(y+3)=0 [-11.66, 33.95, -3.97, 18.85]}