Precalculus Examples

$4 y^{2} - 16 y - x^{2} + 6 x - 9 = 0$

Step 1

Add $9$ to both sides of the equation.

$4 y^{2} - 16 y - x^{2} + 6 x = 9$

Step 2

Complete the square for $4 y^{2} - 16 y$ .

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

$a = 4$

$b = - 16$

$c = 0$

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

$d = \frac{- 16}{2 \cdot 4}$

Simplify the right side.

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Cancel the common factor of $- 16$ and $2$ .

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Factor $2$ out of $- 16$ .

$d = \frac{2 \cdot - 8}{2 \cdot 4}$

Cancel the common factors.

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Factor $2$ out of $2 \cdot 4$ .

$d = \frac{2 \cdot - 8}{2 (4)}$

Cancel the common factor.

$d = \frac{2 \cdot - 8}{2 \cdot 4}$

Rewrite the expression.

$d = \frac{- 8}{4}$

Cancel the common factor of $- 8$ and $4$ .

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Factor $4$ out of $- 8$ .

$d = \frac{4 \cdot - 2}{4}$

Cancel the common factors.

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Factor $4$ out of $4$ .

$d = \frac{4 \cdot - 2}{4 (1)}$

Cancel the common factor.

$d = \frac{4 \cdot - 2}{4 \cdot 1}$

Rewrite the expression.

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

Divide $- 2$ by $1$ .

$d = - 2$

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

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

Simplify the right side.

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Simplify each term.

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Raise $- 16$ to the power of $2$ .

$e = 0 - \frac{256}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{256}{16}$

Divide $256$ by $16$ .

$e = 0 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Subtract $16$ from $0$ .

$e = - 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(y - 2)}^{2} - 16$ .

$4 {(y - 2)}^{2} - 16$

Step 3

Substitute $4 {(y - 2)}^{2} - 16$ for $4 y^{2} - 16 y$ in the equation $4 y^{2} - 16 y - x^{2} + 6 x = 9$ .

$4 {(y - 2)}^{2} - 16 - x^{2} + 6 x = 9$

Step 4

Move $- 16$ to the right side of the equation by adding $16$ to both sides.

$4 {(y - 2)}^{2} - x^{2} + 6 x = 9 + 16$

Step 5

Complete the square for $- x^{2} + 6 x$ .

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

$a = - 1$

$b = 6$

$c = 0$

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

$d = \frac{6}{2 \cdot - 1}$

Simplify the right side.

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Cancel the common factor of $6$ and $2$ .

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Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$d = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$d = - 3$

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

$e = 0 - \frac{6^{2}}{4 \cdot - 1}$

Simplify the right side.

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Simplify each term.

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Raise $6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 0 - \frac{36}{- 4}$

Divide $36$ by $- 4$ .

$e = 0 - - 9$

Multiply $- 1$ by $- 9$ .

$e = 0 + 9$

Add $0$ and $9$ .

$e = 9$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - 3)}^{2} + 9$ .

$- {(x - 3)}^{2} + 9$

Step 6

Substitute $- {(x - 3)}^{2} + 9$ for $- x^{2} + 6 x$ in the equation $4 y^{2} - 16 y - x^{2} + 6 x = 9$ .

$4 {(y - 2)}^{2} - {(x - 3)}^{2} + 9 = 9 + 16$

Step 7

Move $9$ to the right side of the equation by adding $9$ to both sides.

$4 {(y - 2)}^{2} - {(x - 3)}^{2} = 9 + 16 - 9$

Step 8

Simplify $9 + 16 - 9$ .

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Add $9$ and $16$ .

$4 {(y - 2)}^{2} - {(x - 3)}^{2} = 25 - 9$

Subtract $9$ from $25$ .

$4 {(y - 2)}^{2} - {(x - 3)}^{2} = 16$

Step 9

Divide each term by $16$ to make the right side equal to one.

$\frac{4 {(y - 2)}^{2}}{16} - \frac{{(x - 3)}^{2}}{16} = \frac{16}{16}$

Step 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 $1$ .

$\frac{{(y - 2)}^{2}}{4} - \frac{{(x - 3)}^{2}}{16} = 1$