What is the equation of the parabola with a focus at (1,3) and a directrix of y= 2?

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Answer 1 · 2016-04-11T10:11:18

#(x-1)^2=2y-5#

Let their be a point #(x,y)# on parabola. Its distance from focus at #(1,3)# is

#sqrt((x-1)^2+(y-3)^2)#

and its distance from directrix #y=2# will be #y-2#

Hence equation would be

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

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

#(x-1)^2+y^2-6y+9=y^2-4y+4# or

#(x-1)^2=2y-5#

graph{(x-1)^2=2y-5 [-6, 6, -2, 10]}