Trigonometry Examples

$11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Add $889$ to both sides of the equation.

$11 x^{2} - 25 y^{2} + 22 x + 250 y = 889$

Step 1.2

Complete the square for $11 x^{2} + 22 x$ .

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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 $c$ .

$a = 11$

$b = 22$

$c = 0$

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

$d = \frac{22}{2 \cdot 11}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $22$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $22$ .

$d = \frac{2 \cdot 11}{2 \cdot 11}$

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 $2 \cdot 11$ .

$d = \frac{2 \cdot 11}{2 (11)}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 11}{2 \cdot 11}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$d = \frac{11}{11}$

Step 1.2.3.2.2

Cancel the common factor of $11$ .

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Step 1.2.3.2.2.1

Cancel the common factor.

$d = \frac{11}{11}$

Step 1.2.3.2.2.2

Rewrite the expression.

$d = 1$

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

$e = 0 - \frac{22^{2}}{4 \cdot 11}$

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 $22$ to the power of $2$ .

$e = 0 - \frac{484}{4 \cdot 11}$

Step 1.2.4.2.1.2

Multiply $4$ by $11$ .

$e = 0 - \frac{484}{44}$

Step 1.2.4.2.1.3

Divide $484$ by $44$ .

$e = 0 - 1 \cdot 11$

Step 1.2.4.2.1.4

Multiply $- 1$ by $11$ .

$e = 0 - 11$

Step 1.2.4.2.2

Subtract $11$ from $0$ .

$e = - 11$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $11 {(x + 1)}^{2} - 11$ .

$11 {(x + 1)}^{2} - 11$

Step 1.3

Substitute $11 {(x + 1)}^{2} - 11$ for $11 x^{2} + 22 x$ in the equation $11 x^{2} - 25 y^{2} + 22 x + 250 y = 889$ .

$11 {(x + 1)}^{2} - 11 - 25 y^{2} + 250 y = 889$

Step 1.4

Move $- 11$ to the right side of the equation by adding $11$ to both sides.

$11 {(x + 1)}^{2} - 25 y^{2} + 250 y = 889 + 11$

Step 1.5

Complete the square for $- 25 y^{2} + 250 y$ .

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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 $c$ .

$a = - 25$

$b = 250$

$c = 0$

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

$d = \frac{250}{2 \cdot - 25}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

Cancel the common factor of $250$ and $2$ .

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Step 1.5.3.2.1.1

Factor $2$ out of $250$ .

$d = \frac{2 \cdot 125}{2 \cdot - 25}$

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 $2 \cdot - 25$ .

$d = \frac{2 \cdot 125}{2 (- 25)}$

Step 1.5.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 125}{2 \cdot - 25}$

Step 1.5.3.2.1.2.3

Rewrite the expression.

$d = \frac{125}{- 25}$

Step 1.5.3.2.2

Cancel the common factor of $125$ and $- 25$ .

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Step 1.5.3.2.2.1

Factor $25$ out of $125$ .

$d = \frac{25 (5)}{- 25}$

Step 1.5.3.2.2.2

Move the negative one from the denominator of $\frac{5}{- 1}$ .

$d = - 1 \cdot 5$

Step 1.5.3.2.3

Multiply $- 1$ by $5$ .

$d = - 5$

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

$e = 0 - \frac{250^{2}}{4 \cdot - 25}$

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 $250$ to the power of $2$ .

$e = 0 - \frac{62500}{4 \cdot - 25}$

Step 1.5.4.2.1.2

Multiply $4$ by $- 25$ .

$e = 0 - \frac{62500}{- 100}$

Step 1.5.4.2.1.3

Divide $62500$ by $- 100$ .

$e = 0 - - 625$

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 625$ .

$e = 0 + 625$

Step 1.5.4.2.2

Add $0$ and $625$ .

$e = 625$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 25 {(y - 5)}^{2} + 625$ .

$- 25 {(y - 5)}^{2} + 625$

Step 1.6

Substitute $- 25 {(y - 5)}^{2} + 625$ for $- 25 y^{2} + 250 y$ in the equation $11 x^{2} - 25 y^{2} + 22 x + 250 y = 889$ .

$11 {(x + 1)}^{2} - 25 {(y - 5)}^{2} + 625 = 889 + 11$

Step 1.7

Move $625$ to the right side of the equation by adding $625$ to both sides.

$11 {(x + 1)}^{2} - 25 {(y - 5)}^{2} = 889 + 11 - 625$

Step 1.8

Simplify $889 + 11 - 625$ .

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Step 1.8.1

Add $889$ and $11$ .

$11 {(x + 1)}^{2} - 25 {(y - 5)}^{2} = 900 - 625$

Step 1.8.2

Subtract $625$ from $900$ .

$11 {(x + 1)}^{2} - 25 {(y - 5)}^{2} = 275$

Step 1.9

Divide each term by $275$ to make the right side equal to one.

$\frac{11 {(x + 1)}^{2}}{275} - \frac{25 {(y - 5)}^{2}}{275} = \frac{275}{275}$

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 $1$ .

$\frac{{(x + 1)}^{2}}{25} - \frac{{(y - 5)}^{2}}{11} = 1$

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, $a$ .

$a = 5$

$b = \sqrt{11}$

$k = 5$

$h = - 1$

Step 4

Find the vertices.

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Step 4.1

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

$(h + a, k)$

Step 4.2

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

$(4, 5)$

Step 4.3

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

$(h - a, k)$

Step 4.4

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

$(- 6, 5)$

Step 4.5

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

$(4, 5), (- 6, 5)$

Step 5