# How do you identify the foci of this ellipse 25x^2 + 200x + 9y^2 + 54y = -480?

Nov 11, 2015

$\left(- 4 , - \frac{41}{15}\right) , \left(- 4 , - \frac{49}{15}\right)$

#### Explanation:

First, group the $x$'s and the $y$'s.

$25 {x}^{2} + 200 x + 9 {y}^{2} + 54 y = - 480$

$\implies \left(25 {x}^{2} + 200 x\right) + \left(9 {y}^{2} + 54 y\right) = - 480$

$\implies 25 \left({x}^{2} + 8 x\right) + 9 \left({y}^{2} + 6 y\right) = - 480$

Then, we "complete" the square.
To do this, we add something to our binomials such that it will become a perfect square trinomial. However, when we do this, we need to add the same value to the other side of the equation so that the equality is retained

$\implies 25 \left({x}^{2} + 8 x + 16\right) + 9 \left({y}^{2} + 6 y + 9\right) = - 480 + 25 \left(16\right) + 9 \left(9\right)$

$\implies 25 {\left(x + 4\right)}^{2} + 9 {\left(y + 3\right)}^{2} = - 480 + 400 + 81$

$\implies 25 {\left(x + 4\right)}^{2} + 9 {\left(y + 3\right)}^{2} = 1$

Since the other side of the equation is already 1, our equation is already in the form

$\implies {\left(x - h\right)}^{2} / {b}^{2} + {\left(y - k\right)}^{2} / {a}^{2} = 1$

Now let's get the value of $a$ and $b$

$\implies \frac{1}{b} ^ 2 {\left(x - h\right)}^{2} + \frac{1}{a} ^ 2 {\left(y - k\right)}^{2} = 1$

$\implies \frac{1}{b} ^ 2 = 25$
$\implies \frac{1}{25} = {b}^{2}$

$\implies b = \frac{1}{5}$

$\implies \frac{1}{a} ^ 2 = 9$
$\implies \frac{1}{9} = {a}^{2}$

$\implies a = \frac{1}{3}$

Now that we have $a$ and $b$, let's get $c$

${c}^{2} = {a}^{2} - {b}^{2}$

$\implies {c}^{2} = {\left(\frac{1}{3}\right)}^{2} - {\left(\frac{1}{5}\right)}^{2}$

$\implies {c}^{2} = \frac{1}{9} - \frac{1}{25}$

$\implies {c}^{2} = \frac{25 - 9}{225}$

$\implies {c}^{2} = \frac{16}{225}$

$\implies c = \frac{4}{15}$

Now that we have $c$, we can get the foci.
To get the foci, we add/subtract $c$ to/from the coordinate associated with $a$. In this case, $a$ is associated with $y$.

Our ellipse is centered at $\left(h , k\right) = \left(- 4 , - 3\right)$

Our foci are at $\left(h , k + c\right) , \left(h , k - c\right)$

$\implies \left(- 4 , - 3 + \frac{4}{15}\right) , \left(- 4 , - 3 - \frac{4}{15}\right)$

$\implies \left(- 4 , - \frac{41}{15}\right) , \left(- 4 , - \frac{49}{15}\right)$