Algebra Examples

$5 x^{2} + 9 y^{2} + 10 x - 54 y + 41$

Step 1

Find the standard form of the ellipse.

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Step 1.1

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

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Step 1.1.1

Subtract $5 x^{2}$ from both sides of the equation.

$y - 5 x^{2} = 9 y^{2} + 10 x - 54 y + 41$

Step 1.1.2

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

$y - 5 x^{2} - 9 y^{2} = 10 x - 54 y + 41$

Step 1.1.3

Subtract $10 x$ from both sides of the equation.

$y - 5 x^{2} - 9 y^{2} - 10 x = - 54 y + 41$

Step 1.1.4

Add $54 y$ to both sides of the equation.

$y - 5 x^{2} - 9 y^{2} - 10 x + 54 y = 41$

Step 1.1.5

Add $y$ and $54 y$ .

$- 5 x^{2} - 9 y^{2} - 10 x + 55 y = 41$

Step 1.2

Complete the square for $- 5 x^{2} - 10 x$ .

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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 $c$ .

$a = - 5$

$b = - 10$

$c = 0$

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

$d = \frac{- 10}{2 \cdot - 5}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

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

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Step 1.2.3.2.1.1

Factor $2$ out of $- 10$ .

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

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 $2 \cdot - 5$ .

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

Step 1.2.3.2.1.2.2

Cancel the common factor.

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

Step 1.2.3.2.1.2.3

Rewrite the expression.

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

Step 1.2.3.2.2

Cancel the common factor of $- 5$ .

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Step 1.2.3.2.2.1

Cancel the common factor.

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

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

$e = 0 - \frac{{(- 10)}^{2}}{4 \cdot - 5}$

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 $2$ .

$e = 0 - \frac{100}{4 \cdot - 5}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 5$ .

$e = 0 - \frac{100}{- 20}$

Step 1.2.4.2.1.3

Divide $100$ by $- 20$ .

$e = 0 - - 5$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 5$ .

$e = 0 + 5$

Step 1.2.4.2.2

Add $0$ and $5$ .

$e = 5$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 5 {(x + 1)}^{2} + 5$ .

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

Step 1.3

Substitute $- 5 {(x + 1)}^{2} + 5$ for $- 5 x^{2} - 10 x$ in the equation $- 5 x^{2} - 9 y^{2} - 10 x + 55 y = 41$ .

$- 5 {(x + 1)}^{2} + 5 - 9 y^{2} + 55 y = 41$

Step 1.4

Move $5$ to the right side of the equation by adding $5$ to both sides.

$- 5 {(x + 1)}^{2} - 9 y^{2} + 55 y = 41 - 5$

Step 1.5

Complete the square for $- 9 y^{2} + 55 y$ .

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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 $c$ .

$a = - 9$

$b = 55$

$c = 0$

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

$d = \frac{55}{2 \cdot - 9}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

Multiply $2$ by $- 9$ .

$d = \frac{55}{- 18}$

Step 1.5.3.2.2

Move the negative in front of the fraction.

$d = - \frac{55}{18}$

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

$e = 0 - \frac{55^{2}}{4 \cdot - 9}$

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 $55$ to the power of $2$ .

$e = 0 - \frac{3025}{4 \cdot - 9}$

Step 1.5.4.2.1.2

Multiply $4$ by $- 9$ .

$e = 0 - \frac{3025}{- 36}$

Step 1.5.4.2.1.3

Move the negative in front of the fraction.

$e = 0 - - \frac{3025}{36}$

Step 1.5.4.2.1.4

Multiply $- - \frac{3025}{36}$ .

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Step 1.5.4.2.1.4.1

Multiply $- 1$ by $- 1$ .

$e = 0 + 1 (\frac{3025}{36})$

Step 1.5.4.2.1.4.2

Multiply $\frac{3025}{36}$ by $1$ .

$e = 0 + \frac{3025}{36}$

Step 1.5.4.2.2

Add $0$ and $\frac{3025}{36}$ .

$e = \frac{3025}{36}$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 9 {(y - \frac{55}{18})}^{2} + \frac{3025}{36}$ .

$- 9 {(y - \frac{55}{18})}^{2} + \frac{3025}{36}$

Step 1.6

Substitute $- 9 {(y - \frac{55}{18})}^{2} + \frac{3025}{36}$ for $- 9 y^{2} + 55 y$ in the equation $- 5 x^{2} - 9 y^{2} - 10 x + 55 y = 41$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} + \frac{3025}{36} = 41 - 5$

Step 1.7

Move $\frac{3025}{36}$ to the right side of the equation by adding $\frac{3025}{36}$ to both sides.

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = 41 - 5 - \frac{3025}{36}$

Step 1.8

Simplify $41 - 5 - \frac{3025}{36}$ .

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Step 1.8.1

Find the common denominator.

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Step 1.8.1.1

Write $41$ as a fraction with denominator $1$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41}{1} - 5 - \frac{3025}{36}$

Step 1.8.1.2

Multiply $\frac{41}{1}$ by $\frac{36}{36}$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41}{1} \cdot \frac{36}{36} - 5 - \frac{3025}{36}$

Step 1.8.1.3

Multiply $\frac{41}{1}$ by $\frac{36}{36}$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41 \cdot 36}{36} - 5 - \frac{3025}{36}$

Step 1.8.1.4

Write $- 5$ as a fraction with denominator $1$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41 \cdot 36}{36} + \frac{- 5}{1} - \frac{3025}{36}$

Step 1.8.1.5

Multiply $\frac{- 5}{1}$ by $\frac{36}{36}$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41 \cdot 36}{36} + \frac{- 5}{1} \cdot \frac{36}{36} - \frac{3025}{36}$

Step 1.8.1.6

Multiply $\frac{- 5}{1}$ by $\frac{36}{36}$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41 \cdot 36}{36} + \frac{- 5 \cdot 36}{36} - \frac{3025}{36}$

Step 1.8.2

Combine the numerators over the common denominator.

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{41 \cdot 36 - 5 \cdot 36 - 3025}{36}$

Step 1.8.3

Simplify each term.

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Step 1.8.3.1

Multiply $41$ by $36$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{1476 - 5 \cdot 36 - 3025}{36}$

Step 1.8.3.2

Multiply $- 5$ by $36$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{1476 - 180 - 3025}{36}$

Step 1.8.4

Simplify the expression.

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Step 1.8.4.1

Subtract $180$ from $1476$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{1296 - 3025}{36}$

Step 1.8.4.2

Subtract $3025$ from $1296$ .

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = \frac{- 1729}{36}$

Step 1.8.4.3

Move the negative in front of the fraction.

$- 5 {(x + 1)}^{2} - 9 {(y - \frac{55}{18})}^{2} = - \frac{1729}{36}$

Step 1.9

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

$5 {(x + 1)}^{2} + 9 {(y - \frac{55}{18})}^{2} = \frac{1729}{36}$

Step 1.10

Divide each term by $\frac{1729}{36}$ to make the right side equal to one.

$\frac{5 {(x + 1)}^{2}}{\frac{1729}{36}} + \frac{9 {(y - \frac{55}{18})}^{2}}{\frac{1729}{36}} = \frac{\frac{1729}{36}}{\frac{1729}{36}}$

Step 1.11

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 $1$ .

$\frac{{(x + 1)}^{2}}{\frac{1729}{180}} + \frac{{(y - \frac{55}{18})}^{2}}{\frac{1729}{324}} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

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

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = \frac{\sqrt{8645}}{30}$

$b = \frac{\sqrt{1729}}{18}$

$k = \frac{55}{18}$

$h = - 1$

Step 4

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

$(- 1, \frac{55}{18})$

Step 5