Hello, Hockey_Guy14!
Determine an equation for the parabola with focus $\displaystyle F(3,0)$
and directrix $\displaystyle x=7.$
There are two types of parabola:
. . $\displaystyle \begin{array}{cccc}\text{Vertical:} & (x-h)^2 \:=\:4p(y-k) \\ \text{Horizontal:} & (y-k)^2 \:=\:4p(x-h) \end{array}$
where $\displaystyle (h,k)$ is the vertex
and $\displaystyle p$ is the distance from the vertex to the focus.
. . (It is also the distance from the vertex to the directrix.)
Graph the focus and directrix. Code:
* |
| * :
| * :
| F V :
- - + - - o - - * - - : -
| 3 5 :
| * :
| * x=7
* |
The vertex is midway between the focus and the directrix: .$\displaystyle V(5,0)$
. . And we see that: .$\displaystyle p = 2$
The parabola always "bends around" the focus
. . and "away from" the directrix.
This parabola opens to the left.
. Its equation is: .$\displaystyle (y-k)^2 \:=\:-4p(x-h)$
Therefore, the equation is: .$\displaystyle (y-0)^2 \:=\:-4(2)(x-5)\quad\Rightarrow\quad y^2 \:=\:-8(x-5)$