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.
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: .$\displaystyle \frac{x^2}{25} + \frac{y^2}{16} \:=\:1$