# How do you find all the critical points to graph -4x^2 + 9y^2 - 36 = 0 including vertices, foci and asymptotes?

Feb 16, 2018

The center is located at the origin, vertices at $\left(0 , 2\right)$ and $\left(0 , - 2\right)$, and foci at $\left(0 , 3.6\right)$ and $\left(0 , - 3.6\right)$; the slopes of the asymptotes are ${y}^{2} = \pm \frac{2}{3} {x}^{2}$
$G r a p h :$ graph{y^2/4-x^2/9=1 [-10, 10, -5, 5]}

#### Explanation:

We can arrange this equation into $9 {y}^{2} - 4 {x}^{2} = 36$ in order to correct format. I also added $36$ to both sides as this will become necessary down the road. Next, I will divide both sides by $36$: $\frac{9 {y}^{2}}{36} - \frac{4 {x}^{2}}{36} = 1$.
Simplify the remaining equation; ${y}^{2} / 4 - {x}^{2} / 9 = 1$

The next step is to find $a$, $b$, and $c$ (excluding the center; it is at the origin). In order to do so, we have to define where they are:
${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2} = 1$

$a$ is always the root of the first denominator in hyperbolic equations, so it would be $2$.

$b$ would be $3$ ($\sqrt{9} = 3$)

$c$ can be found, only in hyperbolic equations, through Pythagora's Theorem: ${a}^{2} + {b}^{2} = {c}^{2}$
c=~3.6

To find the foci, add and subtract the value of $c$ to the variable being divided by $a$ (which is $y$). This value must be directly above or to the side of the center.
The value of the foci are $\left(0 , 3.6\right)$ and $\left(0 , - 3.6\right)$

To find the vertices, add and subtract your $a$ value to the number being divided by it ($y$, again).
Vertices: $\left(0 , 2\right)$ and $\left(0 , - 2\right)$.

To find the slope depends on the format: if ${y}^{2} / {a}^{2} - {x}^{2} / {b}^{2}$, the slopes of the asymptotes are $\pm \frac{a}{b}$. (If your equation looks like this: ${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$, the slope will be $\pm \frac{b}{a}$)
The slopes of the asymptotes are $\pm \frac{2}{3}$