Precalculus Examples

$4 x^{2} - y^{2} + 16 y - 80 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Add $80$ to both sides of the equation.

$4 x^{2} - y^{2} + 16 y = 80$

Step 1.2

Complete the square for $- y^{2} + 16 y$ .

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Step 1.2.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = 16$

$c = 0$

Step 1.2.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.2.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.2.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{16}{2 \cdot - 1}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $16$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $16$ .

$d = \frac{2 \cdot 8}{2 \cdot - 1}$

Step 1.2.3.2.1.2

Move the negative one from the denominator of $\frac{8}{- 1}$ .

$d = - 1 \cdot 8$

Step 1.2.3.2.2

Multiply $- 1$ by $8$ .

$d = - 8$

Step 1.2.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.2.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{16^{2}}{4 \cdot - 1}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

Raise $16$ to the power of $2$ .

$e = 0 - \frac{256}{4 \cdot - 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 1$ .

$e = 0 - \frac{256}{- 4}$

Step 1.2.4.2.1.3

Divide $256$ by $- 4$ .

$e = 0 - - 64$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 64$ .

$e = 0 + 64$

Step 1.2.4.2.2

Add $0$ and $64$ .

$e = 64$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y - 8)}^{2} + 64$ .

$- {(y - 8)}^{2} + 64$

Step 1.3

Substitute $- {(y - 8)}^{2} + 64$ for $- y^{2} + 16 y$ in the equation $4 x^{2} - y^{2} + 16 y = 80$ .

$4 x^{2} - {(y - 8)}^{2} + 64 = 80$

Step 1.4

Move $64$ to the right side of the equation by adding $64$ to both sides.

$4 x^{2} - {(y - 8)}^{2} = 80 - 64$

Step 1.5

Subtract $64$ from $80$ .

$4 x^{2} - {(y - 8)}^{2} = 16$

Step 1.6

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

$\frac{4 x^{2}}{16} - \frac{{(y - 8)}^{2}}{16} = \frac{16}{16}$

Step 1.7

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 $1$ .

$\frac{x^{2}}{4} - \frac{{(y - 8)}^{2}}{16} = 1$

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.

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

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, $a$ .

$a = 2$

$b = 4$

$k = 8$

$h = 0$

Step 4

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

$(0, 8)$

Step 5