# Conic Section equations

• Jan 23rd 2013, 03:22 PM
tomkoolen
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.

• Jan 23rd 2013, 05:58 PM
Soroban
Re: Conic Section equations
Hello, tomkoolen!

Quote:

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: . $\displaystyle \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \;=\;1$
. . where $\displaystyle (h,k)$ is the center,
. . and $\displaystyle a$ and $\displaystyle b$ are the lengths of the semiaxes.

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

Code:

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

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

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

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