# How do you find the a, b, c for (x + 3)^2/9 + (y – 9)^2/25 =1?

##### 1 Answer
Oct 22, 2015

Find them by a bit of rewriting of the equation and using the formula that $c = \sqrt{\setminus m b \otimes {\left\{s e m i m a j \mathmr{and} a \xi s\right\}}^{2} - \setminus m b \otimes {\left\{s e m i \min \mathmr{and} a \xi s\right\}}^{2}}$. See below for the answers.

#### Explanation:

First, rewrite the equation of this ellipse in the form ${\left(\frac{x - h}{a}\right)}^{2} + {\left(\frac{y - k}{b}\right)}^{2} = 1$:

$\frac{{\left(x + 3\right)}^{2}}{9} + \frac{{\left(y - 9\right)}^{2}}{25} = 1$

$\implies {\left(\frac{x - \left(- 3\right)}{3}\right)}^{2} + {\left(\frac{y - 9}{5}\right)}^{2} = 1$.

Hence $a = 3$ and $b = 5$. Since $5 > 3$, the semimajor axis is $b$ and the semiminor axis is $a$.

Therefore, the distance between the center of the ellipse and the two foci is $c = \sqrt{{5}^{2} - {3}^{2}} = \sqrt{25 - 9} = \sqrt{16} = 4$.

FYI, the center of the ellipse is the point whose rectangular coordinates are $\left(x , y\right) = \left(- 3 , 9\right)$. The four vertices are the points whose rectangular coordinates are $\left(x , y\right) = \left(- 3 , 9 \pm 5\right)$ and $\left(x , y\right) = \left(- 3 \pm 3 , 9\right)$.