Graph 4x^2-9y^2=36

$4 x^{2} - 9 y^{2} = 36$

Step 1

Find the standard form of the hyperbola.

Divide each term by $36$ to make the right side equal to one.

$\frac{4 x^{2}}{36} - \frac{9 y^{2}}{36} = \frac{36}{36}$

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 $1$ .

$\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

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, $a$ .

$a = 3$

$b = 2$

$k = 0$

$h = 0$

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.

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

Simplify.

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

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

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

$\sqrt{9 + 4}$

Add $9$ and $4$ .

$\sqrt{13}$

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.

$(\sqrt{13}, 0)$

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.

$(- \sqrt{13}, 0)$

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

$(\sqrt{13}, 0), (- \sqrt{13}, 0)$

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.

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

Simplify the numerator.

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

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

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

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

Add $9$ and $4$ .

$\frac{\sqrt{13}}{3}$

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.

$\frac{2^{2}}{\sqrt{13}}$

Simplify.

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

$\frac{4}{\sqrt{13}}$

Multiply $\frac{4}{\sqrt{13}}$ by $\frac{\sqrt{13}}{\sqrt{13}}$ .

$\frac{4}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}$

Combine and simplify the denominator.

Multiply $\frac{4}{\sqrt{13}}$ by $\frac{\sqrt{13}}{\sqrt{13}}$ .

$\frac{4 \sqrt{13}}{\sqrt{13} \sqrt{13}}$

Raise $\sqrt{13}$ to the power of $1$ .

$\frac{4 \sqrt{13}}{{\sqrt{13}}^{1} \sqrt{13}}$

Raise $\sqrt{13}$ to the power of $1$ .

$\frac{4 \sqrt{13}}{{\sqrt{13}}^{1} {\sqrt{13}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{4 \sqrt{13}}{{\sqrt{13}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{4 \sqrt{13}}{{\sqrt{13}}^{2}}$

Rewrite ${\sqrt{13}}^{2}$ as $13$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{13}$ as $13^{\frac{1}{2}}$ .

$\frac{4 \sqrt{13}}{{(13^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{4 \sqrt{13}}{13^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\frac{4 \sqrt{13}}{13^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{4 \sqrt{13}}{13^{\frac{2}{2}}}$

Rewrite the expression.

$\frac{4 \sqrt{13}}{13^{1}}$

Evaluate the exponent.

$\frac{4 \sqrt{13}}{13}$

Step 10

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

$y = \pm \frac{2}{3} x + 0$

Step 11

Simplify $\frac{2}{3} x + 0$ .

Add $\frac{2}{3} x$ and $0$ .

$y = \frac{2}{3} x$

Combine $\frac{2}{3}$ and $x$ .

$y = \frac{2 x}{3}$

Step 12

Simplify $- \frac{2}{3} x + 0$ .

Add $- \frac{2}{3} x$ and $0$ .

$y = - \frac{2}{3} x$

Combine $x$ and $\frac{2}{3}$ .

$y = - \frac{x \cdot 2}{3}$

Move $2$ to the left of $x$ .

$y = - \frac{2 x}{3}$

Step 13

This hyperbola has two asymptotes.

$y = \frac{2 x}{3}, y = - \frac{2 x}{3}$

Step 14

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

Center: $(0, 0)$

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

Foci: $(\sqrt{13}, 0), (- \sqrt{13}, 0)$

Eccentricity: $\frac{\sqrt{13}}{3}$

Focal Parameter: $\frac{4 \sqrt{13}}{13}$

Asymptotes: $y = \frac{2 x}{3}$ , $y = - \frac{2 x}{3}$

Step 15

Graph 4x^2-9y^2=36

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