Find the Asymptotes (x^2)/16-(y^2)/49=1

$\frac{x^{2}}{16} - \frac{y^{2}}{49} = 1$

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 $1$ .

$\frac{x^{2}}{16} - \frac{y^{2}}{49} = 1$

Step 2

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

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

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, $a$ .

$a = 4$

$b = 7$

$k = 0$

$h = 0$

Step 4

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

$y = \pm \frac{7}{4} x + 0$

Step 5

Simplify $\frac{7}{4} x + 0$ .

Add $\frac{7}{4} x$ and $0$ .

$y = \frac{7}{4} x$

Combine $\frac{7}{4}$ and $x$ .

$y = \frac{7 x}{4}$

Step 6

Simplify $- \frac{7}{4} x + 0$ .

Add $- \frac{7}{4} x$ and $0$ .

$y = - \frac{7}{4} x$

Combine $x$ and $\frac{7}{4}$ .

$y = - \frac{x \cdot 7}{4}$

Move $7$ to the left of $x$ .

$y = - \frac{7 x}{4}$

Step 7

This hyperbola has two asymptotes.

$y = \frac{7 x}{4}, y = - \frac{7 x}{4}$

Step 8

The asymptotes are $y = \frac{7 x}{4}$ and $y = - \frac{7 x}{4}$ .

Asymptotes: $y = \frac{7 x}{4}, y = - \frac{7 x}{4}$

Step 9

Find the Asymptotes (x^2)/16-(y^2)/49=1

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