Find the Focus y^2-4x+4y-4=0

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

Step 1

Rewrite the equation in vertex form.

Isolate $x$ to the left side of the equation.

Move all terms not containing $x$ to the right side of the equation.

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

$- 4 x + 4 y - 4 = - y^{2}$

Subtract $4 y$ from both sides of the equation.

$- 4 x - 4 = - y^{2} - 4 y$

Add $4$ to both sides of the equation.

$- 4 x = - y^{2} - 4 y + 4$

Divide each term in $- 4 x = - y^{2} - 4 y + 4$ by $- 4$ and simplify.

Divide each term in $- 4 x = - y^{2} - 4 y + 4$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{- y^{2}}{- 4} + \frac{- 4 y}{- 4} + \frac{4}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{- y^{2}}{- 4} + \frac{- 4 y}{- 4} + \frac{4}{- 4}$

Divide $x$ by $1$ .

$x = \frac{- y^{2}}{- 4} + \frac{- 4 y}{- 4} + \frac{4}{- 4}$

Simplify the right side.

Simplify each term.

Dividing two negative values results in a positive value.

$x = \frac{y^{2}}{4} + \frac{- 4 y}{- 4} + \frac{4}{- 4}$

Cancel the common factor of $- 4$ .

Cancel the common factor.

$x = \frac{y^{2}}{4} + \frac{- 4 y}{- 4} + \frac{4}{- 4}$

Divide $y$ by $1$ .

$x = \frac{y^{2}}{4} + y + \frac{4}{- 4}$

Divide $4$ by $- 4$ .

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

Complete the square for $\frac{y^{2}}{4} + y - 1$ .

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

$a = \frac{1}{4}$

$b = 1$

$c = - 1$

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

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

Simplify the right side.

Combine $2$ and $\frac{1}{4}$ .

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

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

$d = 1 \cdot 2$

Multiply $2$ by $1$ .

$d = 2$

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

$e = - 1 - \frac{1^{2}}{4 (\frac{1}{4})}$

Simplify the right side.

Simplify each term.

One to any power is one.

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

Combine $4$ and $\frac{1}{4}$ .

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

Divide $4$ by $4$ .

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

Cancel the common factor of $1$ .

Cancel the common factor.

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

Rewrite the expression.

$e = - 1 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$e = - 1 - 1$

Subtract $1$ from $- 1$ .

$e = - 2$

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

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

Set $x$ equal to the new right side.

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

Step 2

Use the vertex form, $x = a {(y - k)}^{2} + h$ , to determine the values of $a$ , $h$ , and $k$ .

$a = \frac{1}{4}$

$h = - 2$

$k = - 2$

Step 3

Find the vertex $(h, k)$ .

$(- 2, - 2)$

Step 4

Find $p$ , the distance from the vertex to the focus.

Find the distance from the vertex to a focus of the parabola by using the following formula.

$\frac{1}{4 a}$

Substitute the value of $a$ into the formula.

$\frac{1}{4 \cdot \frac{1}{4}}$

Simplify.

Combine $4$ and $\frac{1}{4}$ .

$\frac{1}{\frac{4}{4}}$

Simplify by dividing numbers.

Divide $4$ by $4$ .

$\frac{1}{1}$

Divide $1$ by $1$ .

$1$

Step 5

Find the focus.

The focus of a parabola can be found by adding $p$ to the x-coordinate $h$ if the parabola opens left or right.

$(h + p, k)$

Substitute the known values of $h$ , $p$ , and $k$ into the formula and simplify.

$(- 1, - 2)$

Step 6

Find the Focus y^2-4x+4y-4=0

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