Precalculus Examples

$10 y^{2} - 20 x^{2} + 60 y + 160 x - 225 = 0$

Step 1

Add $225$ to both sides of the equation.

$10 y^{2} - 20 x^{2} + 60 y + 160 x = 225$

Step 2

Complete the square for $10 y^{2} + 60 y$ .

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Step 2.1

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

$a = 10$

$b = 60$

$c = 0$

Step 2.2

Consider the vertex form of a parabola.

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

Step 2.3

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

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Step 2.3.1

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

$d = \frac{60}{2 \cdot 10}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Cancel the common factor of $60$ and $2$ .

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Step 2.3.2.1.1

Factor $2$ out of $60$ .

$d = \frac{2 \cdot 30}{2 \cdot 10}$

Step 2.3.2.1.2

Cancel the common factors.

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Step 2.3.2.1.2.1

Factor $2$ out of $2 \cdot 10$ .

$d = \frac{2 \cdot 30}{2 (10)}$

Step 2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 30}{2 \cdot 10}$

Step 2.3.2.1.2.3

Rewrite the expression.

$d = \frac{30}{10}$

Step 2.3.2.2

Cancel the common factor of $30$ and $10$ .

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Step 2.3.2.2.1

Factor $10$ out of $30$ .

$d = \frac{10 \cdot 3}{10}$

Step 2.3.2.2.2

Cancel the common factors.

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Step 2.3.2.2.2.1

Factor $10$ out of $10$ .

$d = \frac{10 \cdot 3}{10 (1)}$

Step 2.3.2.2.2.2

Cancel the common factor.

$d = \frac{10 \cdot 3}{10 \cdot 1}$

Step 2.3.2.2.2.3

Rewrite the expression.

$d = \frac{3}{1}$

Step 2.3.2.2.2.4

Divide $3$ by $1$ .

$d = 3$

Step 2.4

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

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Step 2.4.1

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

$e = 0 - \frac{60^{2}}{4 \cdot 10}$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Raise $60$ to the power of $2$ .

$e = 0 - \frac{3600}{4 \cdot 10}$

Step 2.4.2.1.2

Multiply $4$ by $10$ .

$e = 0 - \frac{3600}{40}$

Step 2.4.2.1.3

Divide $3600$ by $40$ .

$e = 0 - 1 \cdot 90$

Step 2.4.2.1.4

Multiply $- 1$ by $90$ .

$e = 0 - 90$

Step 2.4.2.2

Subtract $90$ from $0$ .

$e = - 90$

Step 2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $10 {(y + 3)}^{2} - 90$ .

$10 {(y + 3)}^{2} - 90$

Step 3

Substitute $10 {(y + 3)}^{2} - 90$ for $10 y^{2} + 60 y$ in the equation $10 y^{2} - 20 x^{2} + 60 y + 160 x = 225$ .

$10 {(y + 3)}^{2} - 90 - 20 x^{2} + 160 x = 225$

Step 4

Move $- 90$ to the right side of the equation by adding $90$ to both sides.

$10 {(y + 3)}^{2} - 20 x^{2} + 160 x = 225 + 90$

Step 5

Complete the square for $- 20 x^{2} + 160 x$ .

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Step 5.1

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

$a = - 20$

$b = 160$

$c = 0$

Step 5.2

Consider the vertex form of a parabola.

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

Step 5.3

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

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Step 5.3.1

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

$d = \frac{160}{2 \cdot - 20}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Cancel the common factor of $160$ and $2$ .

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Step 5.3.2.1.1

Factor $2$ out of $160$ .

$d = \frac{2 \cdot 80}{2 \cdot - 20}$

Step 5.3.2.1.2

Cancel the common factors.

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Step 5.3.2.1.2.1

Factor $2$ out of $2 \cdot - 20$ .

$d = \frac{2 \cdot 80}{2 (- 20)}$

Step 5.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 80}{2 \cdot - 20}$

Step 5.3.2.1.2.3

Rewrite the expression.

$d = \frac{80}{- 20}$

Step 5.3.2.2

Cancel the common factor of $80$ and $- 20$ .

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Step 5.3.2.2.1

Factor $20$ out of $80$ .

$d = \frac{20 (4)}{- 20}$

Step 5.3.2.2.2

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$d = - 1 \cdot 4$

Step 5.3.2.3

Multiply $- 1$ by $4$ .

$d = - 4$

Step 5.4

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

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Step 5.4.1

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

$e = 0 - \frac{160^{2}}{4 \cdot - 20}$

Step 5.4.2

Simplify the right side.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Raise $160$ to the power of $2$ .

$e = 0 - \frac{25600}{4 \cdot - 20}$

Step 5.4.2.1.2

Multiply $4$ by $- 20$ .

$e = 0 - \frac{25600}{- 80}$

Step 5.4.2.1.3

Divide $25600$ by $- 80$ .

$e = 0 - - 320$

Step 5.4.2.1.4

Multiply $- 1$ by $- 320$ .

$e = 0 + 320$

Step 5.4.2.2

Add $0$ and $320$ .

$e = 320$

Step 5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 20 {(x - 4)}^{2} + 320$ .

$- 20 {(x - 4)}^{2} + 320$

Step 6

Substitute $- 20 {(x - 4)}^{2} + 320$ for $- 20 x^{2} + 160 x$ in the equation $10 y^{2} - 20 x^{2} + 60 y + 160 x = 225$ .

$10 {(y + 3)}^{2} - 20 {(x - 4)}^{2} + 320 = 225 + 90$

Step 7

Move $320$ to the right side of the equation by adding $320$ to both sides.

$10 {(y + 3)}^{2} - 20 {(x - 4)}^{2} = 225 + 90 - 320$

Step 8

Simplify $225 + 90 - 320$ .

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Step 8.1

Add $225$ and $90$ .

$10 {(y + 3)}^{2} - 20 {(x - 4)}^{2} = 315 - 320$

Step 8.2

Subtract $320$ from $315$ .

$10 {(y + 3)}^{2} - 20 {(x - 4)}^{2} = - 5$

Step 9

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

$- 10 {(y + 3)}^{2} + 20 {(x - 4)}^{2} = 5$

Step 10

Divide each term by $5$ to make the right side equal to one.

$- \frac{10 {(y + 3)}^{2}}{5} + \frac{20 {(x - 4)}^{2}}{5} = \frac{5}{5}$

Step 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 $1$ .

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