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

Mar 8, 2017

$y = - \frac{1}{22} {\left(x + 15\right)}^{2} - \frac{27}{2}$

#### 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:

$y = \frac{1}{4 f} {\left(x - h\right)}^{2} + k \text{ [1]}$

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:

$f = \frac{{y}_{\text{focus") - y_("directrix}}}{2}$

$f = \frac{- 19 - - 8}{2}$

$f = - \frac{11}{2}$

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

$k = f + {y}_{\text{directrix}}$

$k = - \frac{11}{2} + - 8$

$k = \frac{- 27}{2}$

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]:

$y = \frac{1}{4 \left(- \frac{11}{2}\right)} {\left(x - - 15\right)}^{2} + \frac{- 27}{2}$

Simplifying a bit:

$y = - \frac{1}{22} {\left(x + 15\right)}^{2} - \frac{27}{2}$

Mar 8, 2017

${x}^{2} + 30 x + 22 y + 522 = 0$

#### 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 $\left({x}_{1} , {y}_{1}\right)$ and x_2,y_2) is given by $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$ and

the distance between point $\left({x}_{1} , {y}_{1}\right)$ and line $a x + b y + c = 0$ is |ax_1+by_1+c|/(sqrt(a^2+b^2).

Now distance of a point $\left(x , y\right)$ on parabola from focus at $\left(- 15 , - 19\right)$ is $\sqrt{{\left(x + 15\right)}^{2} + {\left(y + 19\right)}^{2}}$

and its distance from directrix $y = - 8$ or $y + 8 = 0$ is $| y + 8 \frac{|}{\sqrt{{1}^{2} + {0}^{2}}} = | y + 8 |$

Hence, equation of parabola would be

$\sqrt{{\left(x + 15\right)}^{2} + {\left(y + 19\right)}^{2}} = | y + 8 |$ or

${\left(x + 15\right)}^{2} + {\left(y + 19\right)}^{2} = {\left(y + 8\right)}^{2}$ or

${x}^{2} + 30 x + 225 + {y}^{2} + 38 y + 361 = {y}^{2} + 16 y + 64$ or

${x}^{2} + 30 x + 22 y + 522 = 0$

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