Find the Asymptotes 16x^2-4y^2=64

$16 x^{2} - 4 y^{2} = 64$

Find the standard form of the hyperbola.

Divide each term by $64$ to make the right side equal to one.

$\frac{16 x^{2}}{64} - \frac{4 y^{2}}{64} = \frac{64}{64}$

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 $1$ .

$\frac{x^{2}}{4} - \frac{y^{2}}{16} = 1$

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

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$

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, $a$ .

$a = 2$

$b = 4$

$k = 0$

$h = 0$

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

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

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

$y = 2 x$

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

$y = - 2 x$

This hyperbola has two asymptotes.

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

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

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

Find the Asymptotes 16x^2-4y^2=64

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