What is the equation in standard form of the parabola with a focus at (10,-9) and a directrix of y= -14?

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Answer 1 · 2016-02-08T07:46:07

$y=x^2/10-2x-3/2$

from the given focus $(10, -9)$ and equation of directrix $y=-14$ ,
calculate $p$

$p=1/2(-9--14)=5/2$

calculate the vertex $(h, k)$

$h=10$ and $k=(-9+(-14))/2=-23/2$

Vertex $(h, k)=(10, -23/2)$

Use the vertex form
$(x-h)^2=+4p(y-k)$ positive $4p$ because it opens upward

$(x-10)^2=4*(5/2)(y--23/2)$

$(x-10)^2=10(y+23/2)$

$x^2-20x+100=10y+115$
$x^2-20x-15=10y$

$y=x^2/10-2x-3/2$

the graph of $y=x^2/10-2x-3/2$ and the directrix $y=-14$
graph{(y-x^2/10+2x+3/2)(y+14)=0[-35,35,-25,10]}