# Math Help - Conic Section equations

1. ## Conic Section equations

Hello everyone,

Could someone please tell me how they make equations for the following two conics by using the provided data? I can't seem to find the easy way to do it:

*Parabola with focus (5,-3) and directrix y = 7
*Ellipse with foci (0,4) and (6,4) and the length of the long axe being 14.

2. ## Re: Conic Section equations

Hello, tomkoolen!

Ellipse with foci (0,4) and (6,4) and the length of the long axis is 14.

You are expected to know the general formula for an ellipse: . $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \;=\;1$
. . where $(h,k)$ is the center,
. . and $a$ and $b$ are the lengths of the semiaxes.

Also $c$, the distance from the center to a focus, is given by: $c^2 \:=\:|a^2-b^2|$

Code:
            |
| *   *   *
* |     :       *
*   |     :         *
*    |     :b         *
|     :
*   F1|    C:     F2    *
* - - o - - o - - o - - *
*   (0,4)   :   (6,4)   *
|     :
*    |     :          *
*   |     :         *
------*-+-----:-------*------
| *   *   *
|     : -  a=7  - :
Since the ellipse is "horizontal", $a > b.$

The center is halfway between the foci:. $C(3,4)$
We have: . $c=3,\;a=7$

Hence: . $c^2 \:=\:a^2-b^2 \quad\Rightarrow\quad 3^2 \:=\:7^2 - b^2 \quad\Rightarrow\quad b^2 = 40$

The equation is: . $\frac{(x-3)^2}{49} + \frac{(y-4)^2}{40} \;=\;1$