Exercise 1 Page 650

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.

Horizontal Hyperbola \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1 Vertical Hyperbola \frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1

Consider the given equation.

\frac{x ^{2}}{6 ^{2}} - \frac{y ^{2}}{5 ^{2}} = 1

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

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

Using this information, we can identify that the vertices are

(\pm 6, 0) .

Let's substitute

a = 6

and

b = 5

into the formula for the asymptotes and obtain their equations.

y = \pm \frac{b}{a} x

SubstituteII

$a = 6$ , $b = 5$

y = \pm \frac{5}{6} x

The asymptotes are

y = \pm \frac{5}{6} x,

so their slopes are

\pm \frac{5}{6} .

Now, let's calculate

c,

the absolute value of the nonzero coordinate of the foci. To do so, we will substitute

a = 6

and

b = 5

into

c^{2} = a^{2} + b^{2} .

c^{2} = a^{2} + b^{2}

SubstituteII

$a = 6$ , $b = 5$

c^{2} = 6^{2} + 5^{2}

▼

Solve for

c

CalcPow

Calculate power

c^{2} = 36 + 25

AddTerms

Add terms

c^{2} = 61

SqrtEqn

$LHS = RHS$

c = 61

Note that when solving the above equation, we only needed to consider the principal root because

c

is a positive number. The foci of the hyperbola are

(\pm 61, 0) .

Graph

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

Equation	$\frac{x ^{2}}{6 ^{2}} - \frac{y ^{2}}{5 ^{2}} = 1$
Transverse axis	Horizontal
Vertices	$(\pm 6, 0)$
Foci	$(\pm 61, 0)$
Asymptotes	$y = \pm \frac{5}{6} x$

Finally, we can graph our hyperbola!