What is the vertex form of the equation of the parabola with a focus at (20,29) and a directrix of #y=37 #?

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Answer 1 · 2016-05-30T12:49:06

#y=-1/16(x-20)^2+33#

Let their be a point #(x,y)# on parabola. Its distance from focus at #(20,29)# is

#sqrt((x-20)^2+(y-29)^2)#

and its distance from directrix #y=37# will be #|y-37|#

Hence equation would be

#sqrt((x-20)^2+(y-29)^2)=(y-37)# or

#(x-20)^2+(y-29)^2=(y-37)^2# or

#x^2-40x+400+y^2-58y+841=y^2-74y+1369# or

#x^2-40x+16y-128=0#

or #16y=-x^2+40x+128#

or #y=-1/16(x^2-40x+400)+8+400/16#

or #y=-1/16(x-20)^2+33#

graph{x^2-40x+16y-128=0 [-56.3, 103.7, -35.5, 44.5]}