Question

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

Answer 1

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

Explanation:

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`

Answer 2

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

Explanation:

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]}