# Thread: Need some urgent help please. Thanks

1. ## Need some urgent help please. Thanks

Determine an equation for the parabola with focus (3,0) and directrix x=7. Sketch

Anyone help me out? Much appreciated. Thanks

2. Originally Posted by Hockey_Guy14
Determine an equation for the parabola with focus (3,0) and directrix x=7. Sketch

Anyone help me out? Much appreciated. Thanks
The Focus point F and the Directrix line D are located in such a way that any point P of the parabola is the same distance from the focus point F as it is from the directrix line D.

The vertex of the parabola lies directly between the directrix and the focus point. In our case, it is at: (5, 0)

The distance between the graph and both the focus and directix is always equal, so:

$(x - 3)^2 + y^2 = (x - 7)^2 + (y - y)^2$

$x^2 - 6x + 9 + y^2 = x^2 - 14x + 49$

$y^2 = -8x + 40$

$y = \sqrt{40 - 8x}$

3. Hello, Hockey_Guy14!

Determine an equation for the parabola with focus $F(3,0)$
and directrix $x=7.$

There are two types of parabola:

. . $\begin{array}{cccc}\text{Vertical:} & (x-h)^2 \:=\:4p(y-k) \\ \text{Horizontal:} & (y-k)^2 \:=\:4p(x-h) \end{array}$

where $(h,k)$ is the vertex
and $p$ 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: . $V(5,0)$
. . And we see that: . $p = 2$

The parabola always "bends around" the focus
. . and "away from" the directrix.
This parabola opens to the left.
. Its equation is: . $(y-k)^2 \:=\:-4p(x-h)$

Therefore, the equation is: . $(y-0)^2 \:=\:-4(2)(x-5)\quad\Rightarrow\quad y^2 \:=\:-8(x-5)$