What is the vertex form of the equation of the parabola with a focus at (-3,-9) and a directrix of #y=-10#?

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Answer 1 · 2016-02-11T15:17:57

#(x--3)^2=2(y--19/2)#

The vertex of a parabola is always in between the focus and the directrix

From the given, the directrix is lower than the focus. Therefore the parabola opens upward.

p is 1/2 of the distance from the directrix to the focus

#p=1/2(-9--10)=1/2*1=1/2#

vertex #(h, k)=(-3, (-9+(-10))/2)=(-3, -19/2)

#(x-h)^2=4p(y-k)#

#(x--3)^2=4*(1/2)(y--19/2)#

#(x--3)^2=2(y--19/2)#

see the graph with directrix #y=-10#

graph{((x--3)^2-2(y--19/2))(y+10)=0[-25,25,-13,13]}

have a nice day from the Philippines