Exercise 46 Page 649

We will find the desired information, and use it to draw the graph of the hyperbola.

Vertices, Foci, and Asymptotes

Let's start by recalling the equation of hyperbolas centered at the origin. Now we will rewrite the given equation to match one of these formats.

$$\begin{array}{c}\frac{y^2}{9}-\frac{x^2}{4}=1\iff \frac{y^2}{3^2}-\frac{x^2}{2^2}=1\end{array}$$

From the above formula, we can see that the equation represents a vertical hyperbola. Next, let's review the main characteristics of this type of hyperbola.

Vertical Hyperbola with Center $(0,0)$
Equation	$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Transverse axis	Vertical
Vertices	$(0,\pm a)$
Foci	$(0,\pm c),$ where $c^2=a^2+b^2$
Asymptotes	$y=\pm \frac{a}{b}x$

Using this information, we can identify that the vertices are $\left(0,\pm 3\right).$ Let's substitute $a=3$ and $b=2$ into the formula for the asymptotes and obtain their equations.

$$\begin{array}{c}y=\pm \frac{a}{b}x\Rightarrow y=\pm \frac{3}{2}x\end{array}$$

The asymptotes are $y=\pm \frac{3}{2}x.$ Now, let's calculate $c,$ the absolute value of the nonzero coordinate of the foci. To do so, we will substitute $a=3$ and $b=2$ into $c^2=a^2+b^2.$

$c^2=a^2+b^2$

SubstituteII

$a=3$, $b=2$

$c^2=3^2+2^2$

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Solve for $c$

CalcPow

Calculate power

$c^2=9+4$

AddTerms

Add terms

$c^2=13$

SqrtEqn

$\sqrt{\text{LHS}}=\sqrt{\text{RHS}}$

$c=\pm \sqrt{13}$

CalcRoot

Calculate root

$c=\pm 3.60555\dots$

RoundDec

Round to $2$ decimal place(s)

$c\approx \pm 3.61$

We can now use the formula $(0,\pm c)$ to find the foci of the hyperbola.

$$\begin{array}{c}(0,3.61)\\ (0,\text{-}3.61)\end{array}$$

Graph

To graph the function, let's summarize all of the information that we have found.

Equation	$\frac{y^2}{3^2}-\frac{x^2}{2^2}=1$
Transverse axis	Vertical
Vertices	$\left(0,\pm 3\right)$
Foci	$\left(0,\pm 3.61\right)$
Asymptotes	$y=\pm \frac{3}{2}x$

Finally, we can graph our hyperbola!