Algebra Examples

$(- 4, 8)$ $x = - 6$

Step 1

Since the directrix is horizontal, use the equation of a parabola that opens left or right.

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

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 $x$ coordinate of the vertex using the formula $x = \frac{x coordinate of focus + directrix}{2}$ . The $y$ coordinate will be the same as the $y$ coordinate of the focus.

$(\frac{- 4 - 6}{2}, 8)$

Step 2.2

Simplify the vertex.

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Step 2.2.1

Cancel the common factor of $- 4 - 6$ and $2$ .

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Step 2.2.1.1

Factor $2$ out of $- 4$ .

$(\frac{2 \cdot - 2 - 6}{2}, 8)$

Step 2.2.1.2

Factor $2$ out of $- 6$ .

$(\frac{2 \cdot - 2 + 2 \cdot - 3}{2}, 8)$

Step 2.2.1.3

Factor $2$ out of $2 \cdot - 2 + 2 \cdot - 3$ .

$(\frac{2 \cdot (- 2 - 3)}{2}, 8)$

Step 2.2.1.4

Cancel the common factors.

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Step 2.2.1.4.1

Factor $2$ out of $2$ .

$(\frac{2 \cdot (- 2 - 3)}{2 (1)}, 8)$

Step 2.2.1.4.2

Cancel the common factor.

$(\frac{2 \cdot (- 2 - 3)}{2 \cdot 1}, 8)$

Step 2.2.1.4.3

Rewrite the expression.

$(\frac{- 2 - 3}{1}, 8)$

Step 2.2.1.4.4

Divide $- 2 - 3$ by $1$ .

$(- 2 - 3, 8)$

Step 2.2.2

Subtract $3$ from $- 2$ .

$(- 5, 8)$

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 $x$ coordinate of the vertex from the $x$ coordinate of the focus to find $p$ .

$p = - 4 + 5$

Step 3.2

Add $- 4$ and $5$ .

$p = 1$

Step 4

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

${(y - 8)}^{2} = 4 (1) (x + 5)$

Step 5

Simplify.

${(y - 8)}^{2} = 4 (x + 5)$

Step 6