What is the vertex form of the equation of the parabola with a focus at (1,20) and a directrix of #y=23 #?

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Answer 1 · 2017-10-04T05:54:39

#y=x^2/-6+x/3+64/3#

Given -

Focus #(1,20)#
directrix #y=23#

The vertex of the parabola is in the first quadrant. Its directrix is above the vertex. Hence the parabola opens downward.

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The general form of the equation is -

#(x-h)^2=--4xxaxx(y-k)#

Where -

#h=1# [ X-coordinate of the vertex]
#k=21.5# [ Y-coordinate of the vertex]

Then -

#(x-1)^2=-4xx1.5xx(y-21.5)#

#x^2-2x+1=-6y+129#
#-6y+129=x^2-2x+1#
#-6y=x^2-2x+1-129#
#y=x^2/-6+x/3+128/6#
#y=x^2/-6+x/3+64/3#