What is the equation of the parabola with a focus at (-1,-2) and a directrix of y= -10?

1 Answer
Jan 20, 2016

#y=x^2/16+x/8-95/16#

Explanation:

Let #(x_0, y_0)# be a point on the parabola.
Focus of the parabola is given at #(-1, -2)#
Distance between the two points is
#sqrt((x_0-(-1))^2+(y_0-(-2))^2#
or #sqrt((x_0+1)^2+(y_0+2)^2#
Now distance between the point #(x_0,y_0)# and the given directrix #y=-10 #, is
#|y_0-(-10)|#
#|y_0+10|#

Equate the two distance expressions and squaring both sides.
#(x_0+1)^2+(y_0+2)^2=(y_0+10)^2#
or #(x_0^2+2x_0+1)+(y_0^2+4y_0+4)=(y_0^2+20y_0+100)#

Rearranging and taking term containing #y_0# to one side
#x_0^2+2x_0+1+4-100=20y_0-4y_0#
#y_0=x_0^2/16+x_0/8-95/16#

For any point #(x,y)# this must be true. Therefore, the equation of the parabola is
#y=x^2/16+x/8-95/16#