What is the equation of the parabola with a focus at (1,3) and a directrix of y= 2?

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Answer 1 · 2016-04-11T10:11:18

${(x - 1)}^{2} = 2 y - 5$ $(x-1)^2=2y-5$

Let their be a point $(x, y)$ $(x,y)$ on parabola. Its distance from focus at $(1, 3)$ $(1,3)$ is

$\sqrt{{(x - 1)}^{2} + {(y - 3)}^{2}}$ $sqrt((x-1)^2+(y-3)^2)$

and its distance from directrix $y = 2$ $y=2$ will be $y - 2$ $y-2$

Hence equation would be

$\sqrt{{(x - 1)}^{2} + {(y - 3)}^{2}} = (y - 2)$ $sqrt((x-1)^2+(y-3)^2)=(y-2)$ or

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

${(x - 1)}^{2} + y^{2} - 6 y + 9 = y^{2} - 4 y + 4$ $(x-1)^2+y^2-6y+9=y^2-4y+4$ or

${(x - 1)}^{2} = 2 y - 5$ $(x-1)^2=2y-5$

graph{(x-1)^2=2y-5 [-6, 6, -2, 10]}