Write in Standard Form 16x^2-4y^2=12y-16x+1

$16 x^{2} - 4 y^{2} = 12 y - 16 x + 1$

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

Subtract $12 y$ from both sides of the equation.

$16 x^{2} - 4 y^{2} - 12 y = - 16 x + 1$

Add $16 x$ to both sides of the equation.

$16 x^{2} - 4 y^{2} - 12 y + 16 x = 1$

Move $- 12 y$ .

$16 x^{2} - 4 y^{2} + 16 x - 12 y = 1$

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

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

$a = 16$

$b = 16$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

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

Simplify the right side.

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

Factor $2$ out of $16$ .

$d = \frac{2 \cdot 8}{2 \cdot 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 16$ .

$d = \frac{2 \cdot 8}{2 (16)}$

Cancel the common factor.

$d = \frac{2 \cdot 8}{2 \cdot 16}$

Rewrite the expression.

$d = \frac{8}{16}$

Cancel the common factor of $8$ and $16$ .

Factor $8$ out of $8$ .

$d = \frac{8 (1)}{16}$

Cancel the common factors.

Factor $8$ out of $16$ .

$d = \frac{8 \cdot 1}{8 \cdot 2}$

Cancel the common factor.

$d = \frac{8 \cdot 1}{8 \cdot 2}$

Rewrite the expression.

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

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

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

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $16^{2}$ and $16$ .

Factor $16$ out of $16^{2}$ .

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

Cancel the common factors.

Factor $16$ out of $4 \cdot 16$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $16$ and $4$ .

Factor $4$ out of $16$ .

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

Cancel the common factors.

Factor $4$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $16 {(x + \frac{1}{2})}^{2} - 4$ .

$16 {(x + \frac{1}{2})}^{2} - 4$

Substitute $16 {(x + \frac{1}{2})}^{2} - 4$ for $16 x^{2} + 16 x$ in the equation $16 x^{2} - 4 y^{2} + 16 x - 12 y = 1$ .

$16 {(x + \frac{1}{2})}^{2} - 4 - 4 y^{2} - 12 y = 1$

Move $- 4$ to the right side of the equation by adding $4$ to both sides.

$16 {(x + \frac{1}{2})}^{2} - 4 y^{2} - 12 y = 1 + 4$

Complete the square for $- 4 y^{2} - 12 y$ .

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

$a = - 4$

$b = - 12$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{- 12}{2 \cdot - 4}$

Simplify the right side.

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$d = \frac{2 \cdot - 6}{2 \cdot - 4}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 6}{2 (- 4)}$

Cancel the common factor.

$d = \frac{2 \cdot - 6}{2 \cdot - 4}$

Rewrite the expression.

$d = \frac{- 6}{- 4}$

Cancel the common factor of $- 6$ and $- 4$ .

Factor $- 2$ out of $- 6$ .

$d = \frac{- 2 (3)}{- 4}$

Cancel the common factors.

Factor $- 2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{3}{2}$

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

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

$e = 0 - \frac{{(- 12)}^{2}}{4 \cdot - 4}$

Simplify the right side.

Simplify each term.

Raise $- 12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot - 4}$

Multiply $4$ by $- 4$ .

$e = 0 - \frac{144}{- 16}$

Divide $144$ by $- 16$ .

$e = 0 - - 9$

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 $- 4 {(y + \frac{3}{2})}^{2} + 9$ .

$- 4 {(y + \frac{3}{2})}^{2} + 9$

Substitute $- 4 {(y + \frac{3}{2})}^{2} + 9$ for $- 4 y^{2} - 12 y$ in the equation $16 x^{2} - 4 y^{2} + 16 x - 12 y = 1$ .

$16 {(x + \frac{1}{2})}^{2} - 4 {(y + \frac{3}{2})}^{2} + 9 = 1 + 4$

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

$16 {(x + \frac{1}{2})}^{2} - 4 {(y + \frac{3}{2})}^{2} = 1 + 4 - 9$

Simplify $1 + 4 - 9$ .

Add $1$ and $4$ .

$16 {(x + \frac{1}{2})}^{2} - 4 {(y + \frac{3}{2})}^{2} = 5 - 9$

Subtract $9$ from $5$ .

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

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

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

Divide each term by $4$ to make the right side equal to one.

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

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 $1$ .

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

Write in Standard Form 16x^2-4y^2=12y-16x+1

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