Algebra Examples

$(0, - 2)$ $y = 2$

Step 1

Since the directrix is vertical, use the equation of a parabola that opens up or down.

${(x - h)}^{2} = 4 p (y - k)$

Step 2

Find the vertex.

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Step 2.1

The vertex $(h, k)$ is halfway between the directrix and focus. Find the $y$ coordinate of the vertex using the formula $y = \frac{y coordinate of focus + directrix}{2}$ . The $x$ coordinate will be the same as the $x$ coordinate of the focus.

$(0, \frac{- 2 + 2}{2})$

Step 2.2

Simplify the vertex.

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Step 2.2.1

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

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Step 2.2.1.1

Factor $2$ out of $- 2$ .

$(0, \frac{2 \cdot - 1 + 2}{2})$

Step 2.2.1.2

Factor $2$ out of $2$ .

$(0, \frac{2 \cdot - 1 + 2 \cdot 1}{2})$

Step 2.2.1.3

Factor $2$ out of $2 \cdot - 1 + 2 \cdot 1$ .

$(0, \frac{2 \cdot (- 1 + 1)}{2})$

Step 2.2.1.4

Cancel the common factors.

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Step 2.2.1.4.1

Factor $2$ out of $2$ .

$(0, \frac{2 \cdot (- 1 + 1)}{2 (1)})$

Step 2.2.1.4.2

Cancel the common factor.

$(0, \frac{2 \cdot (- 1 + 1)}{2 \cdot 1})$

Step 2.2.1.4.3

Rewrite the expression.

$(0, \frac{- 1 + 1}{1})$

Step 2.2.1.4.4

Divide $- 1 + 1$ by $1$ .

$(0, - 1 + 1)$

Step 2.2.2

Add $- 1$ and $1$ .

$(0, 0)$

Step 3

Find the distance from the focus to the vertex.

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Step 3.1

The distance from the focus to the vertex and from the vertex to the directrix is $| p |$ . Subtract the $y$ coordinate of the vertex from the $y$ coordinate of the focus to find $p$ .

$p = - 2 - 0$

Step 3.2

Subtract $0$ from $- 2$ .

$p = - 2$

Step 4

Substitute in the known values for the variables into the equation ${(x - h)}^{2} = 4 p (y - k)$ .

${(x - 0)}^{2} = 4 (- 2) (y - 0)$

Step 5

Simplify.

$x^{2} = - 8 y$

Step 6