# More conics help. Urgent

• Jan 7th 2008, 05:48 AM
Hockey_Guy14
More conics help. Urgent
I've tried this question many times and can't seem to get it. Anyone help me out?

Find an equation for the ellipse centered at the origin with one focus (0,-3) and the sum of the focal radii equal to 10. Sketch
• Jan 7th 2008, 07:08 AM
Soroban
Hello, Hockey_Guy14!

If you're having trouble with ellipses,
. . perhaps you don't understand the basic concepts.
But I must admit: this problem can be somewhat tricky.

Quote:

Find an equation for the ellipse centered at the origin
with one focus (0,-3) and the sum of the focal radii equal to 10. Sketch.

An ellipse centered at the origin has the equation: .$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$
where $\displaystyle a$ is the "x-radius" and $\displaystyle b$ is the "y-radius".

$\displaystyle c$ is the "focal distance", distance from the center to a focus.
. . Formula: .$\displaystyle c^2\:=\:|a^2-b^2|$

Since one focus is (0,-3), the other is (0,3) . . . and $\displaystyle c = 3.$

We also know that the ellipse is "horizontal": .$\displaystyle a > b$
Code:

                  b|                   * * *      P             *      |      *         *          |  *  *  *       *            *      *    *                 *  |      *       * - - * - - - + - - - * - - *             F2      |      F1    a

The "focal radii" are the distances $\displaystyle PF_1\text{ and }PF_2.$
. . We are told that: .$\displaystyle PF_1 + PF_2 \:=\:10$

Here's a lesser-used fact: .$\displaystyle PF_1 + PF_2 \:=\:2a$
. . The sum of the focal radii is twice the semimajor axis.

So we have:. . $\displaystyle 2a = 10\quad\Rightarrow\quad a = 5 \quad\Rightarrow\quad\boxed{a^2 = 25}$

From the formula, we have: .$\displaystyle c^2 \:=\:a^2-b^2\quad\Rightarrow\quad 3^2 \:=\:5^2-b^2\quad\Rightarrow\quad\boxed{ b^2 = 16}$

Therefore, the equation is: .$\displaystyle \frac{x^2}{25} + \frac{y^2}{16} \:=\:1$