What is the equation of the parabola with a focus at (-15,-19) and a directrix of y= -8?

Because the directrix is a horizontal line, we know that the parabola is vertically oriented (opens either up or down). Because the y coordinate of the focus (-19) below the directrix (-8), we know that the parabola opens down. The vertex form of the equation for this type of parabola is:

#y = 1/(4f)(x - h)^2+ k" [1]"#

Where h is the x coordinate of the vertex, k it the y coordinated of the vertex, and the focal distance, f, is the half of the signed distance from directrix to the focus:

#f = (y_("focus") - y_("directrix"))/2#

#f = (-19 - -8)/2#

#f = -11/2#

The y coordinate of the vertex, k, is f plus the y coordinate of the directrix:

#k= f +y_("directrix")#

#k = -11/2+ -8#

#k = (-27)/2#

The x coordinate of the vertex, h, is the same as the x coordinate of the focus:

#h = -15#

Substituting these values into equation [1]:

#y = 1/(4(-11/2))(x - -15)^2+ (-27)/2#

Simplifying a bit:

#y = -1/22(x +15)^2- 27/2#

Parabola is the locus of a point, which moves so that its distance from a line, called directix, and a point, called focus, are equal.

We know that the distance between two points #(x_1,y_1)# and #x_2,y_2)# is given by #sqrt((x_2-x_1)^2+(y_2-y_1)^2)# and

the distance between point #(x_1,y_1)# and line #ax+by+c=0# is #|ax_1+by_1+c|/(sqrt(a^2+b^2)#.

Now distance of a point #(x,y)# on parabola from focus at #(-15,-19)# is #sqrt((x+15)^2+(y+19)^2)#

and its distance from directrix #y=-8# or #y+8=0# is #|y+8|/sqrt(1^2+0^2)=|y+8|#

Hence, equation of parabola would be

#sqrt((x+15)^2+(y+19)^2)=|y+8|# or

#(x+15)^2+(y+19)^2=(y+8)^2# or

#x^2+30x+225+y^2+38y+361=y^2+16y+64# or

#x^2+30x+22y+522=0#

graph{x^2+30x+22y+522=0 [-56.5, 23.5, -35.28, 4.72]}