Algebra Examples

$(1, 2)$

$y = - 2$

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.

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

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

$2$ .

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

Step 2.2.1.2

Factor

$2$ out of

$- 2$ .

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

Step 2.2.1.3

Factor

$2$ out of

$2 \cdot 1 + 2 \cdot - 1$ .

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

Step 2.2.1.4

Cancel the common factors.

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Step 2.2.1.4.1

Factor

$2$ out of

$2$ .

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

Step 2.2.1.4.2

Cancel the common factor.

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

Step 2.2.1.4.3

Rewrite the expression.

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

Step 2.2.1.4.4

Divide

$1 - 1$ by

$1$ .

$(1, 1 - 1)$

Step 2.2.2

Subtract

$1$ from

$1$ .

$(1, 0)$

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

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

${(x - 1)}^{2} = 4 (2) (y - 0)$

Step 5

Simplify.

${(x - 1)}^{2} = 8 y$

Step 6