What is the equation, in standard form, for a parabola with the vertex (1,2) and directrix y=-2?

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Answer 1 · 2016-11-29T06:26:26

The equation of the parabola is #(x-1)^2=16(y-2#

The vertex is #(a,b)=(1,2)#

The directrix is #y=-2#

The directrix is also #y=b-p/2#

Therefore,

#-2=2-p/2#

#p/2=4#

#p=8#

The focus is #(a,b+p/2)=(1,2+4)=(1,6)#

#b+p/2=6#

#p/2=6-2=4#

#p=8#

The distance any point #(x,y)# on the parabola is equidisdant from the directrix and the focus.

#y+2=sqrt((x-1)^2+(y-6)^2)#

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

#y^2+4y+4=(x-1)^2+y^2-12y+36#

#16y-32=(x-1)^2#

#(x-1)^2=16(y-2)#

The equation of the parabola is

#(x-1)^2=16(y-2)#

graph{(x-1)^2=16(y-2) [-10, 10, -5, 5]}