What is the equation of a parabola with a focus at #(3, 2)# and a directrix at #y=-4#?

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Answer 1 · 2018-08-04T12:45:41

The equation of the parabola is #y=1/12(x-3)^2-1#

The focus is #F=(3,2)# and the directrix is #y=-4#

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

#y+4=sqrt((x-3)^2+(y-2)^2)#

Squaring both sides

#(y+4)^2=(x-3)^2+(y-2)^2#

#y^2+8y+16=(x-3)^2+y^2-4y+4#

#12(y+1)=(x-3)^2#

#y+1=1/12(x-3)^2#

The equation is #y=1/12(x-3)^2-1#

graph{(y+1-1/12(x-3)^2)(y+4)=0 [-10, 10, -5, 5]}