How do you find the equation in standard form of the parabola whose focus is at (-2, 4) and whose directrix is at x = 4?

1 Answer
Mar 3, 2017

Answer:

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

Explanation:

Parabola is the locus of a point which moves whose distance from a given line called directrix and a point called focus is always constant.

Here we have directrix as #x=4# and focus is #(-2,4)#

Distance of #(x,y)# from #x=4# is #|x-4|#

and distance from focus #(-2,4)# is #sqrt((x+2)^2+(y-4)^2)#. Hence

#sqrt((x+2)^2+(y-4)^2)=|x-4|#

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

or #x^2+4x+4+y^2-8y+16=x^2-8x+16#

or #y^2-8y+4x+4=0#

and for standard form it is #4x=-y^2+8y-4#

or #x=-1/4y^2+2y-1#
graph{x=-1/4y^2+2y-1 [-12.965, 7.035, -1, 9]}