Find the Asymptotes y^2=36+4x^2

$y^{2} = 36 + 4 x^{2}$

Step 1

Find the standard form of the hyperbola.

Move all terms containing variables to the left side of the equation.

Subtract $4 x^{2}$ from both sides of the equation.

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

Reorder $y^{2}$ and $- 4 x^{2}$ .

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

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

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

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 $1$ .

$\frac{y^{2}}{36} - \frac{x^{2}}{9} = 1$

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, $a$ .

$a = 6$

$b = 3$

$k = 0$

$h = 0$

Step 4

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

$y = \pm 2 \cdot x + 0$

Step 5

Add $2 \cdot x$ and $0$ .

$y = 2 x$

Step 6

Add $- 2 \cdot x$ and $0$ .

$y = - 2 x$

Step 7

This hyperbola has two asymptotes.

$y = 2 x, y = - 2 x$

Step 8

The asymptotes are $y = 2 x$ and $y = - 2 x$ .

Asymptotes: $y = 2 x, y = - 2 x$

Step 9

Find the Asymptotes y^2=36+4x^2

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