What is the equation in standard form of the parabola with a focus at (1,4) and a directrix of y= 2?

1 Answer
Dec 6, 2015

#y=1/4x^2-1/2x+13/4#

Explanation:

If #(x,y)# is a point on a parabola then
#color(white)("XXX")#the perpendicular distance from the directrix to #(x,y)#
is equal to
#color(white)("XXX")#the distance from #(x,y)# to the focus.

If the directrix is #y=2#
then
#color(white)("XXX")#the perpendicular distance from the directrix to #(x,y)# is #abs(y-2)#

If the focus is #(1,4)#
then
#color(white)("XXX")#the distance from #(x,y)# to the focus is #sqrt((x-1)^2+(y-4)^2)#
enter image source here
Therefore
#color(white)("XXX")color(green)(abs(y-2)) = sqrt(color(blue)((x-1)^2)+color(red)((y-4)^2))#

#color(white)("XXX")color(green)(y-2)^2) = color(blue)((x-1)^2)+color(red)((y-4)^2)#

#color(white)("XXX")color(green)(cancel(y^2)-4y+4) = color(blue)(x^2-2x+1) + color(red)(cancel(y^2)-8y+16)#

#color(white)("XXX")4y + 4 = x^2-2x+17#

#color(white)("XXX")4y = x^2 -2x +13#

#color(white)("XXX")y = 1/4x^2 -1/2x + 13/4color(white)("XXX")#(standard form)
graph{1/4x^2-1/2x+13/4 [-5.716, 6.77, 0.504, 6.744]}