Precalculus Examples

$- 320 x = - 304 - 16 x^{2} - 81 y^{2}$

Step 1

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

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Step 1.1

Add $16 x^{2}$ to both sides of the equation.

$- 320 x + 16 x^{2} = - 304 - 81 y^{2}$

Step 1.2

Add $81 y^{2}$ to both sides of the equation.

$- 320 x + 16 x^{2} + 81 y^{2} = - 304$

Step 1.3

Move $- 320 x$ .

$16 x^{2} + 81 y^{2} - 320 x = - 304$

Step 2

Complete the square for $16 x^{2} - 320 x$ .

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Step 2.1

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

$a = 16$

$b = - 320$

$c = 0$

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

$d = \frac{- 320}{2 \cdot 16}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Cancel the common factor of $- 320$ and $2$ .

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Step 2.3.2.1.1

Factor $2$ out of $- 320$ .

$d = \frac{2 \cdot - 160}{2 \cdot 16}$

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 $2 \cdot 16$ .

$d = \frac{2 \cdot - 160}{2 (16)}$

Step 2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 160}{2 \cdot 16}$

Step 2.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 160}{16}$

Step 2.3.2.2

Cancel the common factor of $- 160$ and $16$ .

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Step 2.3.2.2.1

Factor $16$ out of $- 160$ .

$d = \frac{16 \cdot - 10}{16}$

Step 2.3.2.2.2

Cancel the common factors.

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Step 2.3.2.2.2.1

Factor $16$ out of $16$ .

$d = \frac{16 \cdot - 10}{16 (1)}$

Step 2.3.2.2.2.2

Cancel the common factor.

$d = \frac{16 \cdot - 10}{16 \cdot 1}$

Step 2.3.2.2.2.3

Rewrite the expression.

$d = \frac{- 10}{1}$

Step 2.3.2.2.2.4

Divide $- 10$ by $1$ .

$d = - 10$

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

$e = 0 - \frac{{(- 320)}^{2}}{4 \cdot 16}$

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 $- 320$ to the power of $2$ .

$e = 0 - \frac{102400}{4 \cdot 16}$

Step 2.4.2.1.2

Multiply $4$ by $16$ .

$e = 0 - \frac{102400}{64}$

Step 2.4.2.1.3

Divide $102400$ by $64$ .

$e = 0 - 1 \cdot 1600$

Step 2.4.2.1.4

Multiply $- 1$ by $1600$ .

$e = 0 - 1600$

Step 2.4.2.2

Subtract $1600$ from $0$ .

$e = - 1600$

Step 2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $16 {(x - 10)}^{2} - 1600$ .

$16 {(x - 10)}^{2} - 1600$

Step 3

Substitute $16 {(x - 10)}^{2} - 1600$ for $16 x^{2} - 320 x$ in the equation $16 x^{2} + 81 y^{2} - 320 x = - 304$ .

$16 {(x - 10)}^{2} - 1600 + 81 y^{2} = - 304$

Step 4

Move $- 1600$ to the right side of the equation by adding $1600$ to both sides.

$16 {(x - 10)}^{2} + 81 y^{2} = - 304 + 1600$

Step 5

Add $- 304$ and $1600$ .

$16 {(x - 10)}^{2} + 81 y^{2} = 1296$

Step 6

Divide each term by $1296$ to make the right side equal to one.

$\frac{16 {(x - 10)}^{2}}{1296} + \frac{81 y^{2}}{1296} = \frac{1296}{1296}$

Step 7

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 $1$ .

$\frac{{(x - 10)}^{2}}{81} + \frac{y^{2}}{16} = 1$