Graph 9y^2-16x^2=144

$9 y^{2} - 16 x^{2} = 144$

Step 1

Find the standard form of the hyperbola.

Divide each term by $144$ to make the right side equal to one.

$\frac{9 y^{2}}{144} - \frac{16 x^{2}}{144} = \frac{144}{144}$

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 $1$ .

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

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.

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

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, $a$ .

$a = 4$

$b = 3$

$k = 0$

$h = 0$

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.

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

Simplify.

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

$\sqrt{16 + {(3)}^{2}}$

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

$\sqrt{16 + 9}$

Add $16$ and $9$ .

$\sqrt{25}$

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 $k$ .

$(h, k + a)$

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

$(0, 4)$

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

$(h, k - a)$

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

$(0, - 4)$

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

$(0, 4), (0, - 4)$

Step 7

Find the foci.

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

$(h, k + c)$

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

$(0, 5)$

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

$(h, k - c)$

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

$(0, - 5)$

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

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

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.

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

Simplify the numerator.

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

$\frac{\sqrt{16 + 3^{2}}}{4}$

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

$\frac{\sqrt{16 + 9}}{4}$

Add $16$ and $9$ .

$\frac{\sqrt{25}}{4}$

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

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

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

$\frac{5}{4}$

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.

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

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

$\frac{9}{5}$

Step 10

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

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

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: $(0, 4), (0, - 4)$

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

Eccentricity: $\frac{5}{4}$

Focal Parameter: $\frac{9}{5}$

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

Step 15

Graph 9y^2-16x^2=144

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