Graph (y^2)/16-(x+4)^2=1

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

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 $1$ .

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

Step 2

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

$a = 4$

$b = 1$

$k = 0$

$h = - 4$

Step 4

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(- 4, 0)$

Step 5

Find $c$ , the distance from the center to a focus.

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(4)}^{2} + {(1)}^{2}}$

Simplify.

Raise $4$ to the power of $2$ .

$\sqrt{16 + {(1)}^{2}}$

One to any power is one.

$\sqrt{16 + 1}$

Add $16$ and $1$ .

$\sqrt{17}$

Step 6

Find the vertices.

The first vertex of a hyperbola can be found by adding $a$ to $k$ .

$(h, k + a)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(- 4, 4)$

The second vertex of a hyperbola can be found by subtracting $a$ from $k$ .

$(h, k - a)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(- 4, - 4)$

The vertices of a hyperbola follow the form of $(h, k \pm a)$ . Hyperbolas have two vertices.

$(- 4, 4), (- 4, - 4)$

Step 7

Find the foci.

The first focus of a hyperbola can be found by adding $c$ to $k$ .

$(h, k + c)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(- 4, \sqrt{17})$

The second focus of a hyperbola can be found by subtracting $c$ from $k$ .

$(h, k - c)$

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(- 4, - \sqrt{17})$

The foci of a hyperbola follow the form of $(h, k \pm \sqrt{a^{2} + b^{2}})$ . Hyperbolas have two foci.

$(- 4, \sqrt{17}), (- 4, - \sqrt{17})$

Step 8

Find the eccentricity.

Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} + b^{2}}}{a}$

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(4)}^{2} + {(1)}^{2}}}{4}$

Simplify the numerator.

Raise $4$ to the power of $2$ .

$\frac{\sqrt{16 + 1^{2}}}{4}$

One to any power is one.

$\frac{\sqrt{16 + 1}}{4}$

Add $16$ and $1$ .

$\frac{\sqrt{17}}{4}$

Step 9

Find the focal parameter.

Find the value of the focal parameter of the hyperbola by using the following formula.

$\frac{b^{2}}{\sqrt{a^{2} + b^{2}}}$

Substitute the values of $b$ and $\sqrt{a^{2} + b^{2}}$ in the formula.

$\frac{1^{2}}{\sqrt{17}}$

Simplify.

One to any power is one.

$\frac{1}{\sqrt{17}}$

Multiply $\frac{1}{\sqrt{17}}$ by $\frac{\sqrt{17}}{\sqrt{17}}$ .

$\frac{1}{\sqrt{17}} \cdot \frac{\sqrt{17}}{\sqrt{17}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{17}}$ by $\frac{\sqrt{17}}{\sqrt{17}}$ .

$\frac{\sqrt{17}}{\sqrt{17} \sqrt{17}}$

Raise $\sqrt{17}$ to the power of $1$ .

$\frac{\sqrt{17}}{{\sqrt{17}}^{1} \sqrt{17}}$

Raise $\sqrt{17}$ to the power of $1$ .

$\frac{\sqrt{17}}{{\sqrt{17}}^{1} {\sqrt{17}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\sqrt{17}}{{\sqrt{17}}^{1 + 1}}$

Add $1$ and $1$ .

$\frac{\sqrt{17}}{{\sqrt{17}}^{2}}$

Rewrite ${\sqrt{17}}^{2}$ as $17$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{17}$ as $17^{\frac{1}{2}}$ .

$\frac{\sqrt{17}}{{(17^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{\sqrt{17}}{17^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\frac{\sqrt{17}}{17^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{\sqrt{17}}{17^{\frac{2}{2}}}$

Rewrite the expression.

$\frac{\sqrt{17}}{17^{1}}$

Evaluate the exponent.

$\frac{\sqrt{17}}{17}$

Step 10

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

$y = \pm 4 \cdot (x - (- 4)) + 0$

Step 11

Simplify to find the first asymptote.

Remove parentheses.

$y = 4 \cdot (x - (- 4)) + 0$

Simplify $4 \cdot (x - (- 4)) + 0$ .

Simplify the expression.

Add $4 \cdot (x - (- 4))$ and $0$ .

$y = 4 \cdot (x - (- 4))$

Multiply $- 1$ by $- 4$ .

$y = 4 \cdot (x + 4)$

Apply the distributive property.

$y = 4 x + 4 \cdot 4$

Multiply $4$ by $4$ .

$y = 4 x + 16$

Step 12

Simplify to find the second asymptote.

Remove parentheses.

$y = - 4 \cdot (x - (- 4)) + 0$

Simplify $- 4 \cdot (x - (- 4)) + 0$ .

Simplify the expression.

Add $- 4 \cdot (x - (- 4))$ and $0$ .

$y = - 4 \cdot (x - (- 4))$

Multiply $- 1$ by $- 4$ .

$y = - 4 \cdot (x + 4)$

Apply the distributive property.

$y = - 4 x - 4 \cdot 4$

Multiply $- 4$ by $4$ .

$y = - 4 x - 16$

Step 13

This hyperbola has two asymptotes.

$y = 4 x + 16, y = - 4 x - 16$

Step 14

These values represent the important values for graphing and analyzing a hyperbola.

Center: $(- 4, 0)$

Vertices: $(- 4, 4), (- 4, - 4)$

Foci: $(- 4, \sqrt{17}), (- 4, - \sqrt{17})$

Eccentricity: $\frac{\sqrt{17}}{4}$

Focal Parameter: $\frac{\sqrt{17}}{17}$

Asymptotes: $y = 4 x + 16$ , $y = - 4 x - 16$

Step 15

Graph (y^2)/16-(x+4)^2=1

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