What is the equation of the parabola with a focus at (0, -1) and a directrix of y = 1?

Question

9926 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-30T13:19:59

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

Let their be a point #(x,y)# on parabola. Its distance from focus at #(0,-1)# is

#sqrt((x-0)^2+(y+1)^2)#

and its distance from directrix #y=1# will be #|y-1|#

Hence equation would be

#sqrt((x-0)^2+(y+1)^2)=(y-1)# or

#(x-0)^2+(y+1)^2=(y-1)^2# or

#x^2+y^2+2y+1=y^2-2y+1# or

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

graph{x^2+2x+4y=0 [-10, 10, -5, 5]}