Find the Asymptotes ((y^2)/16)-((x^2)/9)=1

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

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 $1$ .

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

Step 2

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

$a = 4$

$b = 3$

$k = 0$

$h = 0$

Step 4

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

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

Step 5

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 6

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 7

This hyperbola has two asymptotes.

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

Step 8

The asymptotes are $y = \frac{4 x}{3}$ and $y = - \frac{4 x}{3}$ .

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

Step 9

Find the Asymptotes ((y^2)/16)-((x^2)/9)=1

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