Precalculus Examples

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

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Divide each term by $16$ to make the right side equal to one.

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

Step 1.2

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 $1$ .

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

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, $a$ .

$a = 4$

$b = 4$

$k = 0$

$h = 0$

Step 4

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

$y = \pm 1 \cdot x + 0$

Step 5

Simplify $1 \cdot x + 0$ .

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Step 5.1

Add $1 \cdot x$ and $0$ .

$y = 1 \cdot x$

Step 5.2

Multiply $x$ by $1$ .

$y = x$

Step 6

Simplify $- 1 \cdot x + 0$ .

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Step 6.1

Add $- 1 \cdot x$ and $0$ .

$y = - 1 \cdot x$

Step 6.2

Rewrite $- 1 x$ as $- x$ .

$y = - x$

Step 7

This hyperbola has two asymptotes.

$y = x, y = - x$

Step 8

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

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

Step 9