What is the standard form of the equation of the parabola with a directrix at x=-5 and a focus at (-2,-5)?

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Answer 1 · 2017-01-21T11:35:36

The equation is $(y+5)^2=6(x+7/2)$

Any point $(x,y)$ on the parabola is equidistant from the directrix and the focus.

Therefore,

$x+5=sqrt((x+2)^2+(y+5)^2)$

$(x+5)^2=(x+2)^2+(y+5)^2$

$x^2+10x+25=x^2+4x+4+(y+5)^2$

$(y+5)^2=6x+21$

$(y+5)^2=6(x+7/2)$

The vertex is $(-7/2,-5)$

graph{((y+5)^2-6(x+7/2))(y-100x-500)((x+2)^2+(y+5)^2-0.05)=0 [-28.86, 28.86, -20.2, 8.68]}