Algebra Examples

$9 y^{2} - x^{2} + 2 x + 54 y + 62$

Step 1

Find the standard form of the hyperbola.

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Move all terms containing variables to the left side of the equation.

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Subtract $9 y^{2}$ from both sides of the equation.

$y - 9 y^{2} = - x^{2} + 2 x + 54 y + 62$

Add $x^{2}$ to both sides of the equation.

$y - 9 y^{2} + x^{2} = 2 x + 54 y + 62$

Subtract $2 x$ from both sides of the equation.

$y - 9 y^{2} + x^{2} - 2 x = 54 y + 62$

Subtract $54 y$ from both sides of the equation.

$y - 9 y^{2} + x^{2} - 2 x - 54 y = 62$

Subtract $54 y$ from $y$ .

$- 9 y^{2} + x^{2} - 2 x - 53 y = 62$

Reorder $- 9 y^{2}$ and $x^{2}$ .

$x^{2} - 9 y^{2} - 2 x - 53 y = 62$

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

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

$a = 1$

$b = - 2$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

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

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

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Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ by $1$ .

$d = - 1$

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

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

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

Simplify the right side.

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Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Cancel the common factor of $4$ .

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Cancel the common factor.

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

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Subtract $1$ from $0$ .

$e = - 1$

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

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

Substitute ${(x - 1)}^{2} - 1$ for $x^{2} - 2 x$ in the equation $x^{2} - 9 y^{2} - 2 x - 53 y = 62$ .

${(x - 1)}^{2} - 1 - 9 y^{2} - 53 y = 62$

Move $- 1$ to the right side of the equation by adding $1$ to both sides.

${(x - 1)}^{2} - 9 y^{2} - 53 y = 62 + 1$

Complete the square for $- 9 y^{2} - 53 y$ .

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

$a = - 9$

$b = - 53$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{- 53}{2 \cdot - 9}$

Simplify the right side.

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Multiply $2$ by $- 9$ .

$d = \frac{- 53}{- 18}$

Dividing two negative values results in a positive value.

$d = \frac{53}{18}$

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

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

$e = 0 - \frac{{(- 53)}^{2}}{4 \cdot - 9}$

Simplify the right side.

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Simplify each term.

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Raise $- 53$ to the power of $2$ .

$e = 0 - \frac{2809}{4 \cdot - 9}$

Multiply $4$ by $- 9$ .

$e = 0 - \frac{2809}{- 36}$

Move the negative in front of the fraction.

$e = 0 - - \frac{2809}{36}$

Multiply $- - \frac{2809}{36}$ .

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Multiply $- 1$ by $- 1$ .

$e = 0 + 1 (\frac{2809}{36})$

Multiply $\frac{2809}{36}$ by $1$ .

$e = 0 + \frac{2809}{36}$

Add $0$ and $\frac{2809}{36}$ .

$e = \frac{2809}{36}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 9 {(y + \frac{53}{18})}^{2} + \frac{2809}{36}$ .

$- 9 {(y + \frac{53}{18})}^{2} + \frac{2809}{36}$

Substitute $- 9 {(y + \frac{53}{18})}^{2} + \frac{2809}{36}$ for $- 9 y^{2} - 53 y$ in the equation $x^{2} - 9 y^{2} - 2 x - 53 y = 62$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} + \frac{2809}{36} = 62 + 1$

Move $\frac{2809}{36}$ to the right side of the equation by adding $\frac{2809}{36}$ to both sides.

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = 62 + 1 - \frac{2809}{36}$

Simplify $62 + 1 - \frac{2809}{36}$ .

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Find the common denominator.

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Write $62$ as a fraction with denominator $1$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62}{1} + 1 - \frac{2809}{36}$

Multiply $\frac{62}{1}$ by $\frac{36}{36}$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62}{1} \cdot \frac{36}{36} + 1 - \frac{2809}{36}$

Multiply $\frac{62}{1}$ by $\frac{36}{36}$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62 \cdot 36}{36} + 1 - \frac{2809}{36}$

Write $1$ as a fraction with denominator $1$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62 \cdot 36}{36} + \frac{1}{1} - \frac{2809}{36}$

Multiply $\frac{1}{1}$ by $\frac{36}{36}$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62 \cdot 36}{36} + \frac{1}{1} \cdot \frac{36}{36} - \frac{2809}{36}$

Multiply $\frac{1}{1}$ by $\frac{36}{36}$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62 \cdot 36}{36} + \frac{36}{36} - \frac{2809}{36}$

Combine the numerators over the common denominator.

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{62 \cdot 36 + 36 - 2809}{36}$

Simplify the expression.

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Multiply $62$ by $36$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{2232 + 36 - 2809}{36}$

Add $2232$ and $36$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{2268 - 2809}{36}$

Subtract $2809$ from $2268$ .

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = \frac{- 541}{36}$

Move the negative in front of the fraction.

${(x - 1)}^{2} - 9 {(y + \frac{53}{18})}^{2} = - \frac{541}{36}$

Flip the sign on each term of the equation so the term on the right side is positive.

$- {(x - 1)}^{2} + 9 {(y + \frac{53}{18})}^{2} = \frac{541}{36}$

Divide each term by $\frac{541}{36}$ to make the right side equal to one.

$- \frac{{(x - 1)}^{2}}{\frac{541}{36}} + \frac{9 {(y + \frac{53}{18})}^{2}}{\frac{541}{36}} = \frac{\frac{541}{36}}{\frac{541}{36}}$

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 $1$ .

$\frac{{(y + \frac{53}{18})}^{2}}{\frac{541}{324}} - \frac{{(x - 1)}^{2}}{\frac{541}{36}} = 1$

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, $a$ .

$a = \frac{\sqrt{541}}{18}$

$b = \frac{\sqrt{541}}{6}$

$k = - \frac{53}{18}$

$h = 1$

Step 4

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

$(1, - \frac{53}{18})$

Step 5