What is the equation, in standard form, for a parabola with the vertex (1,2) and directrix y=-2?

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Answer 1 · 2016-11-29T06:26:26

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

The vertex is $(a, b) = (1, 2)$ $(a,b)=(1,2)$

The directrix is $y = - 2$ $y=-2$

The directrix is also $y = b - \frac{p}{2}$ $y=b-p/2$

Therefore,

$- 2 = 2 - \frac{p}{2}$ $-2=2-p/2$

$\frac{p}{2} = 4$ $p/2=4$

$p = 8$ $p=8$

The focus is $(a, b + \frac{p}{2}) = (1, 2 + 4) = (1, 6)$ $(a,b+p/2)=(1,2+4)=(1,6)$

$b + \frac{p}{2} = 6$ $b+p/2=6$

$\frac{p}{2} = 6 - 2 = 4$ $p/2=6-2=4$

$p = 8$ $p=8$

The distance any point $(x, y)$ $(x,y)$ on the parabola is equidisdant from the directrix and the focus.

$y + 2 = \sqrt{{(x - 1)}^{2} + {(y - 6)}^{2}}$ $y+2=sqrt((x-1)^2+(y-6)^2)$

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

$y^{2} + 4 y + 4 = {(x - 1)}^{2} + y^{2} - 12 y + 36$ $y^2+4y+4=(x-1)^2+y^2-12y+36$

$16 y - 32 = {(x - 1)}^{2}$ $16y-32=(x-1)^2$

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

The equation of the parabola is

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

graph{(x-1)^2=16(y-2) [-10, 10, -5, 5]}