How do you find the equation of a parabola with directrix y=0 and focus F(2,2)?

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Answer 1 · 2018-04-02T13:54:11

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

line
#y=0#
representing x axis
point
#F-+(2,2)#

distance from line dL = distance from point dF

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

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

#y^2=x^2-4x+4+y^2-4y+4#

#x^2-4x-4y+4=0#

#x^2-4x+4=4y#

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