What is the equation of the parabola with a focus at (5,2) and a directrix of y= 6?

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Answer 1 · 2016-04-11T10:15:50

#(x-5)^2=-8y+32#

Let their be a point #(x,y)# on parabola. Its distance from focus at #(5,2)# is

#sqrt((x-5)^2+(y-2)^2)#

and its distance from directrix #y=6# will be #y-6#

Hence equation would be

#sqrt((x-5)^2+(y-2)^2)=(y-6)# or

#(x-5)^2+(y-2)^2=(y-6)^2# or

#(x-5)^2+y^2-4y+4=y^2-12y+36# or

#(x-5)^2=-8y+32#

graph{(x-5)^2=-8y+32 [-10, 15, -5, 5]}