Precalculus Examples

$x^{2} - y^{2} - 10 x - 12 y - 12 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Add $12$ to both sides of the equation.

$x^{2} - y^{2} - 10 x - 12 y = 12$

Step 1.2

Complete the square for $x^{2} - 10 x$ .

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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 $c$ .

$a = 1$

$b = - 10$

$c = 0$

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

$d = \frac{- 10}{2 \cdot 1}$

Step 1.2.3.2

Cancel the common factor of $- 10$ and $2$ .

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Step 1.2.3.2.1

Factor $2$ out of $- 10$ .

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Step 1.2.3.2.2

Cancel the common factors.

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Step 1.2.3.2.2.1

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot - 5}{2 (1)}$

Step 1.2.3.2.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Step 1.2.3.2.2.3

Rewrite the expression.

$d = \frac{- 5}{1}$

Step 1.2.3.2.2.4

Divide $- 5$ by $1$ .

$d = - 5$

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

$e = 0 - \frac{{(- 10)}^{2}}{4 \cdot 1}$

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 $- 10$ to the power of $2$ .

$e = 0 - \frac{100}{4 \cdot 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{100}{4}$

Step 1.2.4.2.1.3

Divide $100$ by $4$ .

$e = 0 - 1 \cdot 25$

Step 1.2.4.2.1.4

Multiply $- 1$ by $25$ .

$e = 0 - 25$

Step 1.2.4.2.2

Subtract $25$ from $0$ .

$e = - 25$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 5)}^{2} - 25$ .

${(x - 5)}^{2} - 25$

Step 1.3

Substitute ${(x - 5)}^{2} - 25$ for $x^{2} - 10 x$ in the equation $x^{2} - y^{2} - 10 x - 12 y = 12$ .

${(x - 5)}^{2} - 25 - y^{2} - 12 y = 12$

Step 1.4

Move $- 25$ to the right side of the equation by adding $25$ to both sides.

${(x - 5)}^{2} - y^{2} - 12 y = 12 + 25$

Step 1.5

Complete the square for $- y^{2} - 12 y$ .

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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 $c$ .

$a = - 1$

$b = - 12$

$c = 0$

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

$d = \frac{- 12}{2 \cdot - 1}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

Cancel the common factor of $- 12$ and $2$ .

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Step 1.5.3.2.1.1

Factor $2$ out of $- 12$ .

$d = \frac{2 \cdot - 6}{2 \cdot - 1}$

Step 1.5.3.2.1.2

Move the negative one from the denominator of $\frac{- 6}{- 1}$ .

$d = - 1 \cdot - 6$

Step 1.5.3.2.2

Rewrite $- 1 \cdot - 6$ as $- - 6$ .

$d = - - 6$

Step 1.5.3.2.3

Multiply $- 1$ by $- 6$ .

$d = 6$

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

$e = 0 - \frac{{(- 12)}^{2}}{4 \cdot - 1}$

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

Raise $- 12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot - 1}$

Step 1.5.4.2.1.2

Multiply $4$ by $- 1$ .

$e = 0 - \frac{144}{- 4}$

Step 1.5.4.2.1.3

Divide $144$ by $- 4$ .

$e = 0 - - 36$

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 36$ .

$e = 0 + 36$

Step 1.5.4.2.2

Add $0$ and $36$ .

$e = 36$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 6)}^{2} + 36$ .

$- {(y + 6)}^{2} + 36$

Step 1.6

Substitute $- {(y + 6)}^{2} + 36$ for $- y^{2} - 12 y$ in the equation $x^{2} - y^{2} - 10 x - 12 y = 12$ .

${(x - 5)}^{2} - {(y + 6)}^{2} + 36 = 12 + 25$

Step 1.7

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

${(x - 5)}^{2} - {(y + 6)}^{2} = 12 + 25 - 36$

Step 1.8

Simplify $12 + 25 - 36$ .

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Step 1.8.1

Add $12$ and $25$ .

${(x - 5)}^{2} - {(y + 6)}^{2} = 37 - 36$

Step 1.8.2

Subtract $36$ from $37$ .

${(x - 5)}^{2} - {(y + 6)}^{2} = 1$

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.

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

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, $a$ .

$a = 1$

$b = 1$

$k = - 6$

$h = 5$

Step 4

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

$(5, - 6)$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Simplify.

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Step 5.3.1

One to any power is one.

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

Step 5.3.2

One to any power is one.

$\sqrt{1 + 1}$

Step 5.3.3

Add $1$ and $1$ .

$\sqrt{2}$

Step 6

Find the vertices.

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Step 6.1

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

$(h + a, k)$

Step 6.2

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

$(6, - 6)$

Step 6.3

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

$(h - a, k)$

Step 6.4

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

$(4, - 6)$

Step 6.5

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

$(6, - 6), (4, - 6)$

Step 7

Find the foci.

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Step 7.1

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

$(h + c, k)$

Step 7.2

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

$(5 + \sqrt{2}, - 6)$

Step 7.3

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

$(h - c, k)$

Step 7.4

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

$(5 - \sqrt{2}, - 6)$

Step 7.5

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

$(5 + \sqrt{2}, - 6), (5 - \sqrt{2}, - 6)$

Step 8

Find the eccentricity.

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Step 8.1

Find the eccentricity by using the following formula.

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

Step 8.2

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

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

Step 8.3

Simplify.

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Step 8.3.1

Divide $\sqrt{{(1)}^{2} + {(1)}^{2}}$ by $1$ .

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

Step 8.3.2

One to any power is one.

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

Step 8.3.3

One to any power is one.

$\sqrt{1 + 1}$

Step 8.3.4

Add $1$ and $1$ .

$\sqrt{2}$

Step 9

Find the focal parameter.

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Step 9.1

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

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

Step 9.2

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

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

Step 9.3

Simplify.

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Step 9.3.1

One to any power is one.

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

Step 9.3.2

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Step 9.3.3

Combine and simplify the denominator.

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Step 9.3.3.1

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 9.3.3.2

Raise $\sqrt{2}$ to the power of $1$ .

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

Step 9.3.3.3

Raise $\sqrt{2}$ to the power of $1$ .

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

Step 9.3.3.4

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

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

Step 9.3.3.5

Add $1$ and $1$ .

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

Step 9.3.3.6

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

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Step 9.3.3.6.1

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

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

Step 9.3.3.6.2

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

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

Step 9.3.3.6.3

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

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

Step 9.3.3.6.4

Cancel the common factor of $2$ .

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Step 9.3.3.6.4.1

Cancel the common factor.

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

Step 9.3.3.6.4.2

Rewrite the expression.

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

Step 9.3.3.6.5

Evaluate the exponent.

$\frac{\sqrt{2}}{2}$

Step 10

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

$y = \pm 1 \cdot (x - (5)) - 6$

Step 11

Simplify to find the first asymptote.

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Step 11.1

Remove parentheses.

$y = 1 \cdot (x - (5)) - 6$

Step 11.2

Simplify $1 \cdot (x - (5)) - 6$ .

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Multiply $x - (5)$ by $1$ .

$y = x - (5) - 6$

Step 11.2.1.2

Multiply $- 1$ by $5$ .

$y = x - 5 - 6$

Step 11.2.2

Subtract $6$ from $- 5$ .

$y = x - 11$

Step 12

Simplify to find the second asymptote.

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Step 12.1

Remove parentheses.

$y = - 1 \cdot (x - (5)) - 6$

Step 12.2

Simplify $- 1 \cdot (x - (5)) - 6$ .

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Multiply $- 1$ by $5$ .

$y = - 1 \cdot (x - 5) - 6$

Step 12.2.1.2

Apply the distributive property.

$y = - 1 x - 1 \cdot - 5 - 6$

Step 12.2.1.3

Rewrite $- 1 x$ as $- x$ .

$y = - x - 1 \cdot - 5 - 6$

Step 12.2.1.4

Multiply $- 1$ by $- 5$ .

$y = - x + 5 - 6$

Step 12.2.2

Subtract $6$ from $5$ .

$y = - x - 1$

Step 13

This hyperbola has two asymptotes.

$y = x - 11, y = - x - 1$

Step 14

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

Center: $(5, - 6)$

Vertices: $(6, - 6), (4, - 6)$

Foci: $(5 + \sqrt{2}, - 6), (5 - \sqrt{2}, - 6)$

Eccentricity: $\sqrt{2}$

Focal Parameter: $\frac{\sqrt{2}}{2}$

Asymptotes: $y = x - 11$ , $y = - x - 1$

Step 15