Find the Directrix (x-2)^2=-2(y-5)

${(x - 2)}^{2} = - 2 (y - 5)$

Step 1

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

Rewrite the equation as $- 2 (y - 5) = {(x - 2)}^{2}$ .

$- 2 (y - 5) = {(x - 2)}^{2}$

Divide each term in $- 2 (y - 5) = {(x - 2)}^{2}$ by $- 2$ and simplify.

Divide each term in $- 2 (y - 5) = {(x - 2)}^{2}$ by $- 2$ .

$\frac{- 2 (y - 5)}{- 2} = \frac{{(x - 2)}^{2}}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 (y - 5)}{- 2} = \frac{{(x - 2)}^{2}}{- 2}$

Divide $y - 5$ by $1$ .

$y - 5 = \frac{{(x - 2)}^{2}}{- 2}$

Simplify the right side.

Move the negative in front of the fraction.

$y - 5 = - \frac{{(x - 2)}^{2}}{2}$

Add $5$ to both sides of the equation.

$y = - \frac{{(x - 2)}^{2}}{2} + 5$

Reorder terms.

$y = - \frac{1}{2} \cdot {(x - 2)}^{2} + 5$

Step 2

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

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

$h = 2$

$k = 5$

Step 3

Find the vertex $(h, k)$ .

$(2, 5)$

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.

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

Simplify.

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

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

Move the negative in front of the fraction.

$- \frac{1}{4 (\frac{1}{2})}$

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

$- \frac{1}{\frac{4}{2}}$

Divide $4$ by $2$ .

$- \frac{1}{2}$

Step 5

Find the directrix.

The directrix of a parabola is the horizontal line found by subtracting $p$ from the y-coordinate $k$ of the vertex if the parabola opens up or down.

$y = k - p$

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

$y = \frac{11}{2}$

Step 6

Find the Directrix (x-2)^2=-2(y-5)

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