Precalculus Examples

$16 x^{2} - 9 y^{2} - 64 x - 18 y - 89 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Add $89$ to both sides of the equation.

$16 x^{2} - 9 y^{2} - 64 x - 18 y = 89$

Step 1.2

Complete the square for $16 x^{2} - 64 x$ .

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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 $c$ .

$a = 16$

$b = - 64$

$c = 0$

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

$d = \frac{- 64}{2 \cdot 16}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $- 64$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $- 64$ .

$d = \frac{2 \cdot - 32}{2 \cdot 16}$

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 $2 \cdot 16$ .

$d = \frac{2 \cdot - 32}{2 (16)}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 32}{2 \cdot 16}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 32}{16}$

Step 1.2.3.2.2

Cancel the common factor of $- 32$ and $16$ .

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Step 1.2.3.2.2.1

Factor $16$ out of $- 32$ .

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

Step 1.2.3.2.2.2

Cancel the common factors.

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Step 1.2.3.2.2.2.1

Factor $16$ out of $16$ .

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

Step 1.2.3.2.2.2.2

Cancel the common factor.

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

Step 1.2.3.2.2.2.3

Rewrite the expression.

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

Step 1.2.3.2.2.2.4

Divide $- 2$ by $1$ .

$d = - 2$

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

$e = 0 - \frac{{(- 64)}^{2}}{4 \cdot 16}$

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 $- 64$ to the power of $2$ .

$e = 0 - \frac{4096}{4 \cdot 16}$

Step 1.2.4.2.1.2

Multiply $4$ by $16$ .

$e = 0 - \frac{4096}{64}$

Step 1.2.4.2.1.3

Divide $4096$ by $64$ .

$e = 0 - 1 \cdot 64$

Step 1.2.4.2.1.4

Multiply $- 1$ by $64$ .

$e = 0 - 64$

Step 1.2.4.2.2

Subtract $64$ from $0$ .

$e = - 64$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $16 {(x - 2)}^{2} - 64$ .

$16 {(x - 2)}^{2} - 64$

Step 1.3

Substitute $16 {(x - 2)}^{2} - 64$ for $16 x^{2} - 64 x$ in the equation $16 x^{2} - 9 y^{2} - 64 x - 18 y = 89$ .

$16 {(x - 2)}^{2} - 64 - 9 y^{2} - 18 y = 89$

Step 1.4

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

$16 {(x - 2)}^{2} - 9 y^{2} - 18 y = 89 + 64$

Step 1.5

Complete the square for $- 9 y^{2} - 18 y$ .

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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 $c$ .

$a = - 9$

$b = - 18$

$c = 0$

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

$d = \frac{- 18}{2 \cdot - 9}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

Cancel the common factor of $- 18$ and $2$ .

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Step 1.5.3.2.1.1

Factor $2$ out of $- 18$ .

$d = \frac{2 \cdot - 9}{2 \cdot - 9}$

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 $2 \cdot - 9$ .

$d = \frac{2 \cdot - 9}{2 (- 9)}$

Step 1.5.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 9}{2 \cdot - 9}$

Step 1.5.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 9}{- 9}$

Step 1.5.3.2.2

Cancel the common factor of $- 9$ .

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Step 1.5.3.2.2.1

Cancel the common factor.

$d = \frac{- 9}{- 9}$

Step 1.5.3.2.2.2

Rewrite the expression.

$d = 1$

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

$e = 0 - \frac{{(- 18)}^{2}}{4 \cdot - 9}$

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 $- 18$ to the power of $2$ .

$e = 0 - \frac{324}{4 \cdot - 9}$

Step 1.5.4.2.1.2

Multiply $4$ by $- 9$ .

$e = 0 - \frac{324}{- 36}$

Step 1.5.4.2.1.3

Divide $324$ by $- 36$ .

$e = 0 - - 9$

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 9$ .

$e = 0 + 9$

Step 1.5.4.2.2

Add $0$ and $9$ .

$e = 9$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 9 {(y + 1)}^{2} + 9$ .

$- 9 {(y + 1)}^{2} + 9$

Step 1.6

Substitute $- 9 {(y + 1)}^{2} + 9$ for $- 9 y^{2} - 18 y$ in the equation $16 x^{2} - 9 y^{2} - 64 x - 18 y = 89$ .

$16 {(x - 2)}^{2} - 9 {(y + 1)}^{2} + 9 = 89 + 64$

Step 1.7

Move $9$ to the right side of the equation by adding $9$ to both sides.

$16 {(x - 2)}^{2} - 9 {(y + 1)}^{2} = 89 + 64 - 9$

Step 1.8

Simplify $89 + 64 - 9$ .

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Step 1.8.1

Add $89$ and $64$ .

$16 {(x - 2)}^{2} - 9 {(y + 1)}^{2} = 153 - 9$

Step 1.8.2

Subtract $9$ from $153$ .

$16 {(x - 2)}^{2} - 9 {(y + 1)}^{2} = 144$

Step 1.9

Divide each term by $144$ to make the right side equal to one.

$\frac{16 {(x - 2)}^{2}}{144} - \frac{9 {(y + 1)}^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{{(x - 2)}^{2}}{9} - \frac{{(y + 1)}^{2}}{16} = 1$

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, $a$ .

$a = 3$

$b = 4$

$k = - 1$

$h = 2$

Step 4

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

$(2, - 1)$

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.

$\sqrt{{(3)}^{2} + {(4)}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

Raise $3$ to the power of $2$ .

$\sqrt{9 + {(4)}^{2}}$

Step 5.3.2

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

$\sqrt{9 + 16}$

Step 5.3.3

Add $9$ and $16$ .

$\sqrt{25}$

Step 5.3.4

Rewrite $25$ as $5^{2}$ .

$\sqrt{5^{2}}$

Step 5.3.5

Pull terms out from under the radical, assuming positive real numbers.

$5$

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.

$(5, - 1)$

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.

$(- 1, - 1)$

Step 6.5

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

$(5, - 1), (- 1, - 1)$

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.

$(7, - 1)$

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.

$(- 3, - 1)$

Step 7.5

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

$(7, - 1), (- 3, - 1)$

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.

$\frac{\sqrt{{(3)}^{2} + {(4)}^{2}}}{3}$

Step 8.3

Simplify the numerator.

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Step 8.3.1

Raise $3$ to the power of $2$ .

$\frac{\sqrt{9 + 4^{2}}}{3}$

Step 8.3.2

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

$\frac{\sqrt{9 + 16}}{3}$

Step 8.3.3

Add $9$ and $16$ .

$\frac{\sqrt{25}}{3}$

Step 8.3.4

Rewrite $25$ as $5^{2}$ .

$\frac{\sqrt{5^{2}}}{3}$

Step 8.3.5

Pull terms out from under the radical, assuming positive real numbers.

$\frac{5}{3}$

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.

$\frac{4^{2}}{5}$

Step 9.3

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

$\frac{16}{5}$

Step 10

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

$y = \pm \frac{4}{3} \cdot (x - (2)) - 1$

Step 11

Simplify to find the first asymptote.

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Step 11.1

Remove parentheses.

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

Step 11.2

Simplify $\frac{4}{3} \cdot (x - (2)) - 1$ .

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Multiply $- 1$ by $2$ .

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

Step 11.2.1.2

Apply the distributive property.

$y = \frac{4}{3} x + \frac{4}{3} \cdot - 2 - 1$

Step 11.2.1.3

Combine $\frac{4}{3}$ and $x$ .

$y = \frac{4 x}{3} + \frac{4}{3} \cdot - 2 - 1$

Step 11.2.1.4

Multiply $\frac{4}{3} \cdot - 2$ .

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Step 11.2.1.4.1

Combine $\frac{4}{3}$ and $- 2$ .

$y = \frac{4 x}{3} + \frac{4 \cdot - 2}{3} - 1$

Step 11.2.1.4.2

Multiply $4$ by $- 2$ .

$y = \frac{4 x}{3} + \frac{- 8}{3} - 1$

Step 11.2.1.5

Move the negative in front of the fraction.

$y = \frac{4 x}{3} - \frac{8}{3} - 1$

Step 11.2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{4 x}{3} - \frac{8}{3} - 1 \cdot \frac{3}{3}$

Step 11.2.3

Combine $- 1$ and $\frac{3}{3}$ .

$y = \frac{4 x}{3} - \frac{8}{3} + \frac{- 1 \cdot 3}{3}$

Step 11.2.4

Combine the numerators over the common denominator.

$y = \frac{4 x}{3} + \frac{- 8 - 1 \cdot 3}{3}$

Step 11.2.5

Simplify the numerator.

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Step 11.2.5.1

Multiply $- 1$ by $3$ .

$y = \frac{4 x}{3} + \frac{- 8 - 3}{3}$

Step 11.2.5.2

Subtract $3$ from $- 8$ .

$y = \frac{4 x}{3} + \frac{- 11}{3}$

Step 11.2.6

Move the negative in front of the fraction.

$y = \frac{4 x}{3} - \frac{11}{3}$

Step 12

Simplify to find the second asymptote.

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Step 12.1

Remove parentheses.

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

Step 12.2

Simplify $- \frac{4}{3} \cdot (x - (2)) - 1$ .

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Multiply $- 1$ by $2$ .

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

Step 12.2.1.2

Apply the distributive property.

$y = - \frac{4}{3} x - \frac{4}{3} \cdot - 2 - 1$

Step 12.2.1.3

Combine $x$ and $\frac{4}{3}$ .

$y = - \frac{x \cdot 4}{3} - \frac{4}{3} \cdot - 2 - 1$

Step 12.2.1.4

Multiply $- \frac{4}{3} \cdot - 2$ .

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Step 12.2.1.4.1

Multiply $- 2$ by $- 1$ .

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

Step 12.2.1.4.2

Combine $2$ and $\frac{4}{3}$ .

$y = - \frac{x \cdot 4}{3} + \frac{2 \cdot 4}{3} - 1$

Step 12.2.1.4.3

Multiply $2$ by $4$ .

$y = - \frac{x \cdot 4}{3} + \frac{8}{3} - 1$

Step 12.2.1.5

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{3} + \frac{8}{3} - 1$

Step 12.2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = - \frac{4 x}{3} + \frac{8}{3} - 1 \cdot \frac{3}{3}$

Step 12.2.3

Combine $- 1$ and $\frac{3}{3}$ .

$y = - \frac{4 x}{3} + \frac{8}{3} + \frac{- 1 \cdot 3}{3}$

Step 12.2.4

Combine the numerators over the common denominator.

$y = - \frac{4 x}{3} + \frac{8 - 1 \cdot 3}{3}$

Step 12.2.5

Simplify the numerator.

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Step 12.2.5.1

Multiply $- 1$ by $3$ .

$y = - \frac{4 x}{3} + \frac{8 - 3}{3}$

Step 12.2.5.2

Subtract $3$ from $8$ .

$y = - \frac{4 x}{3} + \frac{5}{3}$

Step 13

This hyperbola has two asymptotes.

$y = \frac{4 x}{3} - \frac{11}{3}, y = - \frac{4 x}{3} + \frac{5}{3}$

Step 14

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

Center: $(2, - 1)$

Vertices: $(5, - 1), (- 1, - 1)$

Foci: $(7, - 1), (- 3, - 1)$

Eccentricity: $\frac{5}{3}$

Focal Parameter: $\frac{16}{5}$

Asymptotes: $y = \frac{4 x}{3} - \frac{11}{3}$ , $y = - \frac{4 x}{3} + \frac{5}{3}$

Step 15