Find the Asymptotes 9y^2-4x^2=36

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

Step 1

Find the standard form of the hyperbola.

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

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

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 $1$ .

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

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find the 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, $a$ .

$a = 2$

$b = 3$

$k = 0$

$h = 0$

Step 4

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

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

Step 5

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 6

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 7

This hyperbola has two asymptotes.

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

Step 8

The asymptotes are $y = \frac{2 x}{3}$ and $y = - \frac{2 x}{3}$ .

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

Step 9

Find the Asymptotes 9y^2-4x^2=36

Related Questions

Support LCMGCF.com by disabling your adblocker.