What is the equation in standard form of the parabola with a focus at (12,5) and a directrix of y= 16?

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Answer 1 · 2016-05-30T12:53:06

$x^{2} - 24 x + 32 y - 87 = 0$ $x^2-24x+32y-87=0$

Let their be a point $(x, y)$ $(x,y)$ on parabola. Its distance from focus at $(12, 5)$ $(12,5)$ is

$\sqrt{{(x - 12)}^{2} + {(y - 5)}^{2}}$ $sqrt((x-12)^2+(y-5)^2)$

and its distance from directrix $y = 16$ $y=16$ will be $| y - 16 |$ $|y-16|$

Hence equation would be

$\sqrt{{(x - 12)}^{2} + {(y - 5)}^{2}} = (y - 16)$ $sqrt((x-12)^2+(y-5)^2)=(y-16)$ or

${(x - 12)}^{2} + {(y - 5)}^{2} = {(y - 16)}^{2}$ $(x-12)^2+(y-5)^2=(y-16)^2$ or

$x^{2} - 24 x + 144 + y^{2} - 10 y + 25 = y^{2} - 32 y + 256$ $x^2-24x+144+y^2-10y+25=y^2-32y+256$ or

$x^{2} - 24 x + 22 y - 87 = 0$ $x^2-24x+22y-87=0$

graph{x^2-24x+22y-87=0 [-27.5, 52.5, -19.84, 20.16]}