# How do you find the center, foci and vertices of 9x^2+4y^2+36x-24y+36=0?

Dec 30, 2016

$C : \left(- 2 , 3\right)$
$V : \left(- 2 , 3 \pm 3\right)$
$f : \left(- 2 , 3 \pm \sqrt{5}\right)$

#### Explanation:

First off, group terms with the same variables together

$9 {x}^{2} + 4 {y}^{2} + 36 x - 24 y + 36 = 0$

$\implies \left(9 {x}^{2} + 36 x\right) + \left(4 {y}^{2} - 24 y\right) + 36 = 0$

$\implies 9 \left({x}^{2} + 4 x\right) + 4 \left({y}^{2} - 6 y\right) + 36 = 0$

Next, we "complete" the square.
Add a third term to each of the grouped terms such that it will be a perfect square trinomial.

${\left(x + a\right)}^{2} = {x}^{2} + 2 a + {a}^{2}$

However, note that the equality must be maintained. To do this, we need to either add the same value on the other side of the equation or subtract the same value on the same side of the equation.

$\implies 9 \left({x}^{2} + 4 x + 4\right) + 4 \left({y}^{2} - 6 y + 9\right) + 36 - 9 \left(4\right) - 4 \left(9\right) = 0$

$\implies 9 {\left(x + 2\right)}^{2} + 4 {\left(y - 3\right)}^{2} + 36 - 36 - 36 = 0$

$\implies 9 {\left(x + 2\right)}^{2} + 4 {\left(y - 3\right)}^{2} - 36 = 0$

Move the constant to the other side

$\implies 9 {\left(x + 2\right)}^{2} + 4 {\left(y - 3\right)}^{2} = 36$

We want the other side of the equation to be 1, so divide both sides of the equation by the moved constant

$\implies \frac{9 {\left(x + 2\right)}^{2} + 4 {\left(y - 3\right)}^{2} = 36}{36}$

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

Now that we have transform the equation into the standard equation for ellipse, we can easily get the desired properties

$C : \left(- 2 , 3\right)$

Major axis is vertical

$\implies V : \left(- 2 , 3 \pm 3\right)$

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

$\implies 9 - 4 = {c}^{2}$

$\implies \sqrt{5} = c$

$\implies f : \left(- 2 , 3 \pm \sqrt{5}\right)$

Dec 30, 2016

Rearrarange the equation thus:
$9 {\left(x + 2\right)}^{2} - 36 + 4 {\left(y - 3\right)}^{2} - 36 + 36 = 0$

#### Explanation:

Then re-arrange to standard form:
${\left(\frac{x + 2}{2}\right)}^{2} + {\left(\frac{y - 3}{3}\right)}^{2} = 1$
which is an ellipse centered at $\left(- 2 , 3\right)$, vertices $\left(0 , 3\right)$, $\left(- 4 , 3\right)$,$\left(- 2 , 0\right)$, $\left(- 2 , 6\right)$. Its foci are possibly $\left(- 2 , 3 \pm \sqrt{5}\right)$.