Precalculus Examples

$4 x^{2} - y^{2} + 24 x + 4 y + 28 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Subtract $28$ from both sides of the equation.

$4 x^{2} - y^{2} + 24 x + 4 y = - 28$

Step 1.2

Complete the square for $4 x^{2} + 24 x$ .

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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 $c$ .

$a = 4$

$b = 24$

$c = 0$

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

$d = \frac{24}{2 \cdot 4}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $24$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $24$ .

$d = \frac{2 \cdot 12}{2 \cdot 4}$

Step 1.2.3.2.1.2

Cancel the common factors.

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Step 1.2.3.2.1.2.1

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

$d = \frac{2 \cdot 12}{2 (4)}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 12}{2 \cdot 4}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$d = \frac{12}{4}$

Step 1.2.3.2.2

Cancel the common factor of $12$ and $4$ .

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Step 1.2.3.2.2.1

Factor $4$ out of $12$ .

$d = \frac{4 \cdot 3}{4}$

Step 1.2.3.2.2.2

Cancel the common factors.

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Step 1.2.3.2.2.2.1

Factor $4$ out of $4$ .

$d = \frac{4 \cdot 3}{4 (1)}$

Step 1.2.3.2.2.2.2

Cancel the common factor.

$d = \frac{4 \cdot 3}{4 \cdot 1}$

Step 1.2.3.2.2.2.3

Rewrite the expression.

$d = \frac{3}{1}$

Step 1.2.3.2.2.2.4

Divide $3$ by $1$ .

$d = 3$

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

$e = 0 - \frac{24^{2}}{4 \cdot 4}$

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 $24$ to the power of $2$ .

$e = 0 - \frac{576}{4 \cdot 4}$

Step 1.2.4.2.1.2

Multiply $4$ by $4$ .

$e = 0 - \frac{576}{16}$

Step 1.2.4.2.1.3

Divide $576$ by $16$ .

$e = 0 - 1 \cdot 36$

Step 1.2.4.2.1.4

Multiply $- 1$ by $36$ .

$e = 0 - 36$

Step 1.2.4.2.2

Subtract $36$ from $0$ .

$e = - 36$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x + 3)}^{2} - 36$ .

$4 {(x + 3)}^{2} - 36$

Step 1.3

Substitute $4 {(x + 3)}^{2} - 36$ for $4 x^{2} + 24 x$ in the equation $4 x^{2} - y^{2} + 24 x + 4 y = - 28$ .

$4 {(x + 3)}^{2} - 36 - y^{2} + 4 y = - 28$

Step 1.4

Move $- 36$ to the right side of the equation by adding $36$ to both sides.

$4 {(x + 3)}^{2} - y^{2} + 4 y = - 28 + 36$

Step 1.5

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

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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 $c$ .

$a = - 1$

$b = 4$

$c = 0$

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

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

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

Cancel the common factor of $4$ and $2$ .

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Step 1.5.3.2.1.1

Factor $2$ out of $4$ .

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

Step 1.5.3.2.1.2

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$d = - 1 \cdot 2$

Step 1.5.3.2.2

Multiply $- 1$ by $2$ .

$d = - 2$

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

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

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

Cancel the common factor of $4^{2}$ and $4$ .

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Step 1.5.4.2.1.1.1

Factor $4$ out of $4^{2}$ .

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

Step 1.5.4.2.1.1.2

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

$e = 0 - (- 1 \cdot 4)$

Step 1.5.4.2.1.2

Multiply $- (- 1 \cdot 4)$ .

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Step 1.5.4.2.1.2.1

Multiply $- 1$ by $4$ .

$e = 0 - - 4$

Step 1.5.4.2.1.2.2

Multiply $- 1$ by $- 4$ .

$e = 0 + 4$

Step 1.5.4.2.2

Add $0$ and $4$ .

$e = 4$

Step 1.5.5

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

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

Step 1.6

Substitute $- {(y - 2)}^{2} + 4$ for $- y^{2} + 4 y$ in the equation $4 x^{2} - y^{2} + 24 x + 4 y = - 28$ .

$4 {(x + 3)}^{2} - {(y - 2)}^{2} + 4 = - 28 + 36$

Step 1.7

Move $4$ to the right side of the equation by adding $4$ to both sides.

$4 {(x + 3)}^{2} - {(y - 2)}^{2} = - 28 + 36 - 4$

Step 1.8

Simplify $- 28 + 36 - 4$ .

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Step 1.8.1

Add $- 28$ and $36$ .

$4 {(x + 3)}^{2} - {(y - 2)}^{2} = 8 - 4$

Step 1.8.2

Subtract $4$ from $8$ .

$4 {(x + 3)}^{2} - {(y - 2)}^{2} = 4$

Step 1.9

Divide each term by $4$ to make the right side equal to one.

$\frac{4 {(x + 3)}^{2}}{4} - \frac{{(y - 2)}^{2}}{4} = \frac{4}{4}$

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 $1$ .

${(x + 3)}^{2} - \frac{{(y - 2)}^{2}}{4} = 1$

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, $a$ .

$a = 1$

$b = 2$

$k = 2$

$h = - 3$

Step 4

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

$(- 3, 2)$

Step 5