What is the equation in standard form of the parabola with a focus at (3,6) and a directrix of y= 7?

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Answer 1 · 2018-01-29T20:48:32

The equation is #y=-1/2(x-3)^2+13/2#

A point on the parabola is equidistant from the directrix and the focus.

The focus is #F=(3,6)#

The directrix is #y=7#

#sqrt((x-3)^2+(y-6)^2)=7-y#

Squaring both sides

#(sqrt((x-3)^2+(y-6)^2))^2=(7-y)^2#

#(x-3)^2+(y-6)^2=(7-y)^2#

#(x-3)^2+y^2-12y+36=49-14y+y^2#

#14y-12y-49=(x-3)^2#

#2y=-(x-3)^2+13#

#y=-1/2(x-3)^2+13/2#

graph{((x-3)^2+2y-13)(y-7)((x-3)^2+(y-6)^2-0.01)=0 [-2.31, 8.79, 3.47, 9.02]}