Precalculus Examples

$- x^{2} + 4 y^{2} + 8 x + 16 y - 64 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Add $64$ to both sides of the equation.

$- x^{2} + 4 y^{2} + 8 x + 16 y = 64$

Step 1.2

Complete the square for $- x^{2} + 8 x$ .

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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 $c$ .

$a = - 1$

$b = 8$

$c = 0$

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 $d = \frac{b}{2 a}$ .

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

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $8$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $8$ .

$d = \frac{2 \cdot 4}{2 \cdot - 1}$

Step 1.2.3.2.1.2

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$d = - 1 \cdot 4$

Step 1.2.3.2.2

Multiply $- 1$ by $4$ .

$d = - 4$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{8^{2}}{4 \cdot - 1}$

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 $8$ to the power of $2$ .

$e = 0 - \frac{64}{4 \cdot - 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 1$ .

$e = 0 - \frac{64}{- 4}$

Step 1.2.4.2.1.3

Divide $64$ by $- 4$ .

$e = 0 - - 16$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 16$ .

$e = 0 + 16$

Step 1.2.4.2.2

Add $0$ and $16$ .

$e = 16$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 4)}^{2} + 16$ .

$- {(x - 4)}^{2} + 16$

Step 1.3

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

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

Step 1.4

Move $16$ to the right side of the equation by adding $16$ to both sides.

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

Step 1.5

Complete the square for $4 y^{2} + 16 y$ .

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Step 1.5.1

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

$a = 4$

$b = 16$

$c = 0$

Step 1.5.2

Consider the vertex form of a parabola.

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

Step 1.5.3

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

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Step 1.5.3.1

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

$d = \frac{16}{2 \cdot 4}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

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

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Step 1.5.3.2.1.1

Factor $2$ out of $16$ .

$d = \frac{2 \cdot 8}{2 \cdot 4}$

Step 1.5.3.2.1.2

Cancel the common factors.

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Step 1.5.3.2.1.2.1

Factor $2$ out of $2 \cdot 4$ .

$d = \frac{2 \cdot 8}{2 (4)}$

Step 1.5.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 8}{2 \cdot 4}$

Step 1.5.3.2.1.2.3

Rewrite the expression.

$d = \frac{8}{4}$

Step 1.5.3.2.2

Cancel the common factor of $8$ and $4$ .

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Step 1.5.3.2.2.1

Factor $4$ out of $8$ .

$d = \frac{4 \cdot 2}{4}$

Step 1.5.3.2.2.2

Cancel the common factors.

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Step 1.5.3.2.2.2.1

Factor $4$ out of $4$ .

$d = \frac{4 \cdot 2}{4 (1)}$

Step 1.5.3.2.2.2.2

Cancel the common factor.

$d = \frac{4 \cdot 2}{4 \cdot 1}$

Step 1.5.3.2.2.2.3

Rewrite the expression.

$d = \frac{2}{1}$

Step 1.5.3.2.2.2.4

Divide $2$ by $1$ .

$d = 2$

Step 1.5.4

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

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Step 1.5.4.1

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

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

Step 1.5.4.2

Simplify the right side.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

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

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

Step 1.5.4.2.1.2

Multiply $4$ by $4$ .

$e = 0 - \frac{256}{16}$

Step 1.5.4.2.1.3

Divide $256$ by $16$ .

$e = 0 - 1 \cdot 16$

Step 1.5.4.2.1.4

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Step 1.5.4.2.2

Subtract $16$ from $0$ .

$e = - 16$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(y + 2)}^{2} - 16$ .

$4 {(y + 2)}^{2} - 16$

Step 1.6

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

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

Step 1.7

Move $- 16$ to the right side of the equation by adding $16$ to both sides.

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

Step 1.8

Simplify $64 - 16 + 16$ .

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Step 1.8.1

Subtract $16$ from $64$ .

$- {(x - 4)}^{2} + 4 {(y + 2)}^{2} = 48 + 16$

Step 1.8.2

Add $48$ and $16$ .

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

Step 1.9

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

$- \frac{{(x - 4)}^{2}}{64} + \frac{4 {(y + 2)}^{2}}{64} = \frac{64}{64}$

Step 1.10

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 $1$ .

$\frac{{(y + 2)}^{2}}{16} - \frac{{(x - 4)}^{2}}{64} = 1$

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, $a$ .

$a = 4$

$b = 8$

$k = - 2$

$h = 4$

Step 4

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

$(4, - 2)$

Step 5