Find the Focus (x+2)^2=-(y+1)

${(x + 2)}^{2} = - (y + 1)$

Step 1

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

Rewrite the equation as $- (y + 1) = {(x + 2)}^{2}$ .

$- (y + 1) = {(x + 2)}^{2}$

Simplify $- (y + 1)$ .

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

$- y - 1 = {(x + 2)}^{2}$

Add $1$ to both sides of the equation.

$- y = {(x + 2)}^{2} + 1$

Divide each term in $- y = {(x + 2)}^{2} + 1$ by $- 1$ and simplify.

Divide each term in $- y = {(x + 2)}^{2} + 1$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{{(x + 2)}^{2}}{- 1}$ .

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

Rewrite $- 1 \cdot {(x + 2)}^{2}$ as $- {(x + 2)}^{2}$ .

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

Divide $1$ by $- 1$ .

$y = - {(x + 2)}^{2} - 1$

Step 2

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

$a = - 1$

$h = - 2$

$k = - 1$

Step 3

Find the vertex $(h, k)$ .

$(- 2, - 1)$

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.

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

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

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

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

Move the negative in front of the fraction.

$- \frac{1}{4}$

Step 5

Find the focus.

The focus of a parabola can be found by adding $p$ to the y-coordinate $k$ if the parabola opens up or down.

$(h, k + p)$

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

$(- 2, - \frac{5}{4})$

Step 6

Find the Focus (x+2)^2=-(y+1)

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