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

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

The directrix is $y=-2$

The directrix is also $y=b-p/2$

Therefore,

$-2=2-p/2$

$p/2=4$

$p=8$

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

$b+p/2=6$

$p/2=6-2=4$

$p=8$

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

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

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

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

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

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

The equation of the parabola is

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

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