Graph (x^2)/9-(y^2)/16=1

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

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

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 3$

$b = 4$

$k = 0$

$h = 0$

Step 4

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(0, 0)$

Step 5

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(3)}^{2} + {(4)}^{2}}$

Simplify.

Raise $3$ to the power of $2$ .

$\sqrt{9 + {(4)}^{2}}$

Raise $4$ to the power of $2$ .

$\sqrt{9 + 16}$

Add $9$ and $16$ .

$\sqrt{25}$

Rewrite $25$ as $5^{2}$ .

$\sqrt{5^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$5$

Step 6

Find the vertices.

The first vertex of a hyperbola can be found by adding $a$ to $h$ .

$(h + a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(3, 0)$

The second vertex of a hyperbola can be found by subtracting $a$ from $h$ .

$(h - a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(- 3, 0)$

The vertices of a hyperbola follow the form of $(h \pm a, k)$ . Hyperbolas have two vertices.

$(3, 0), (- 3, 0)$

Step 7

Find the foci.

The first focus of a hyperbola can be found by adding $c$ to $h$ .

$(h + c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(5, 0)$

The second focus of a hyperbola can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(- 5, 0)$

The foci of a hyperbola follow the form of $(h \pm \sqrt{a^{2} + b^{2}}, k)$ . Hyperbolas have two foci.

$(5, 0), (- 5, 0)$

Step 8

Find the eccentricity.

Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} + b^{2}}}{a}$

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(3)}^{2} + {(4)}^{2}}}{3}$

Simplify the numerator.

Raise $3$ to the power of $2$ .

$\frac{\sqrt{9 + 4^{2}}}{3}$

Raise $4$ to the power of $2$ .

$\frac{\sqrt{9 + 16}}{3}$

Add $9$ and $16$ .

$\frac{\sqrt{25}}{3}$

Rewrite $25$ as $5^{2}$ .

$\frac{\sqrt{5^{2}}}{3}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{5}{3}$

Step 9

Find the focal parameter.

Find the value of the focal parameter of the hyperbola by using the following formula.

$\frac{b^{2}}{\sqrt{a^{2} + b^{2}}}$

Substitute the values of $b$ and $\sqrt{a^{2} + b^{2}}$ in the formula.

$\frac{4^{2}}{5}$

Raise $4$ to the power of $2$ .

$\frac{16}{5}$

Step 10

The asymptotes follow the form $y = \pm \frac{b (x - h)}{a} + k$ because this hyperbola opens left and right.

$y = \pm \frac{4}{3} x + 0$

Step 11

Simplify $\frac{4}{3} x + 0$ .

Add $\frac{4}{3} x$ and $0$ .

$y = \frac{4}{3} x$

Combine $\frac{4}{3}$ and $x$ .

$y = \frac{4 x}{3}$

Step 12

Simplify $- \frac{4}{3} x + 0$ .

Add $- \frac{4}{3} x$ and $0$ .

$y = - \frac{4}{3} x$

Combine $x$ and $\frac{4}{3}$ .

$y = - \frac{x \cdot 4}{3}$

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{3}$

Step 13

This hyperbola has two asymptotes.

$y = \frac{4 x}{3}, y = - \frac{4 x}{3}$

Step 14

These values represent the important values for graphing and analyzing a hyperbola.

Center: $(0, 0)$

Vertices: $(3, 0), (- 3, 0)$

Foci: $(5, 0), (- 5, 0)$

Eccentricity: $\frac{5}{3}$

Focal Parameter: $\frac{16}{5}$

Asymptotes: $y = \frac{4 x}{3}$ , $y = - \frac{4 x}{3}$

Step 15

Graph (x^2)/9-(y^2)/16=1

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