Graph Parabolas Given the Vertex, Focus and Directrix 1

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Question 1 of 4

1. Question
Write the equation of the parabola given the following:

Focus $(0, 3)$ $(0,3)$

Vertex $(0, 0)$ $(0,0)$
- $x^{2} =$ $x^2=$ (12) $y$ $y$
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Standard Form (Concave Up)

$x^{2} = 4$ $x^2=4$ $a$ $a$ $y$ $y$

Focus

$(0,$ $(0,$ $a$ $a$ $)$ $)$

Standard Form (Concave Down)

$x^{2} = - 4$ $x^2=-4$ $a$ $a$ $y$ $y$

Focus

$(0, -$ $(0,-$ $a$ $a$ $)$ $)$

First, plot the given focus and vertex

Focus $(0, 3)$ $(0,3)$

Vertex $(0, 0)$ $(0,0)$

Based on the diagram, the focal length, $a$ $a$ , is equal to $3$ $3$

Now that we know $a$ $a$ , we can also plot the directrix.

$y$ $y$ $=$ $=$ $-$ $-$ $a$ $a$ Directrix Formula

$y$ $y$ $=$ $=$ $-$ $-$ $3$ $3$

Next, draw the parabola. Remember that the parabola’s concavity should be opposite the directrix

Since the parabola is concave up, use $x^{2} = 4$ $x^2=4$ $a$ $a$ $y$ $y$

Finally, form the equation by substituting $a = 3$ $a=3$ to the chosen standard form

$x^{2}$ $x^2$ $=$ $=$ $4$ $4$ $a$ $a$ $y$ $y$ Standard Form

$x^{2}$ $x^2$ $=$ $=$ $4$ $4$ $(3)$ $(3)$ $y$ $y$ Substitute $a = 3$ $a=3$

$x^{2}$ $x^2$ $=$ $=$ $12 y$ $12y$

$x^{2} = 12 y$ $x^2=12y$
Question 2 of 4

2. Question
Write the equation of the parabola given the following:

Focus $(0, - 1)$ $(0,-1)$

Vertex $(0, 0)$ $(0,0)$
- $x^{2} =$ $x^2=$ (-4) $y$ $y$
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Standard Form (Concave Up)

$x^{2} = 4$ $x^2=4$ $a$ $a$ $y$ $y$

Focus

$(0,$ $(0,$ $a$ $a$ $)$ $)$

Standard Form (Concave Down)

$x^{2} = - 4$ $x^2=-4$ $a$ $a$ $y$ $y$

Focus

$(0, -$ $(0,-$ $a$ $a$ $)$ $)$

First, plot the given focus and vertex

Focus $(0, - 1)$ $(0,-1)$

Vertex $(0, 0)$ $(0,0)$

Based on the diagram, the focal length, $a$ $a$ , is equal to $1$ $1$

Now that we know $a$ $a$ , we can also plot the directrix.

$y$ $y$ $=$ $=$ $a$ $a$ Directrix Formula

$y$ $y$ $=$ $=$ $1$ $1$

Next, draw the parabola. Remember that the parabola’s concavity should be opposite the directrix

Since the parabola is concave down, use $x^{2} = - 4$ $x^2=-4$ $a$ $a$ $y$ $y$

Finally, form the equation by substituting $a = 1$ $a=1$ to the chosen standard form

$x^{2}$ $x^2$ $=$ $=$ $- 4$ $-4$ $a$ $a$ $y$ $y$ Standard Form

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

$x^{2}$ $x^2$ $=$ $=$ $- 4 y$ $-4y$

$x^{2} = - 4 y$ $x^2=-4y$
Question 3 of 4

3. Question
Write the equation of the parabola given the following:

Directrix $x = - \frac{1}{4}$ $x=-1/4$

Vertex $(0, 0)$ $(0,0)$
- $y^{2} =$ $y^2=$ (x)
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Standard Form (Concave Right)

$y^{2} = 4$ $y^2=4$ $a$ $a$ $x$ $x$

Focus

$($ $($ $a$ $a$ $, 0)$ $,0)$

Standard Form (Concave Left)

$y^{2} = - 4$ $y^2=-4$ $a$ $a$ $x$ $x$

Focus

$(-$ $(-$ $a$ $a$ $, 0)$ $,0)$

First, plot the given directrix and vertex

Directrix $x = - \frac{1}{4}$ $x=-1/4$

Vertex $(0, 0)$ $(0,0)$

Based on the diagram, the focal length, $a$ $a$ , is equal to $\frac{1}{4}$ $1/4$

Now that we know $a$ $a$ , we can also plot the focus.

$($ $($ $a$ $a$ $, 0)$ $,0)$ becomes $($ $($ $\frac{1}{4}$ $1/4$ $, 0)$ $,0)$

Next, draw the parabola. Remember that the parabola’s concavity should be opposite the directrix

Since the parabola is concave right, use $y^{2} = 4$ $y^2=4$ $a$ $a$ $x$ $x$

Finally, form the equation by substituting $a = \frac{1}{4}$ $a=1/4$ to the chosen standard form

$y^{2}$ $y^2$ $=$ $=$ $4$ $4$ $a$ $a$ $x$ $x$ Standard Form

$y^{2}$ $y^2$ $=$ $=$ $4$ $4$ $(\frac{1}{4})$ $(1/4)$ $x$ $x$ Substitute $a = \frac{1}{4}$ $a=1/4$

$y^{2}$ $y^2$ $=$ $=$ $x$ $x$

$y^{2} = x$ $y^2=x$
Question 4 of 4

4. Question
Write the equation of the parabola given the following:

Directrix $x = 2$ $x=2$

Vertex $(0, 0)$ $(0,0)$
- $y^{2} =$ $y^2=$ (-8) $x$ $x$
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Standard Form (Concave Right)

$y^{2} = 4$ $y^2=4$ $a$ $a$ $x$ $x$

Focus

$($ $($ $a$ $a$ $, 0)$ $,0)$

Standard Form (Concave Left)

$y^{2} = - 4$ $y^2=-4$ $a$ $a$ $x$ $x$

Focus

$(-$ $(-$ $a$ $a$ $, 0)$ $,0)$

First, plot the given directrix and vertex

Directrix $x = 2$ $x=2$

Vertex $(0, 0)$ $(0,0)$

Based on the diagram, the focal length, $a$ $a$ , is equal to $2$ $2$

Now that we know $a$ $a$ , we can also plot the focus.

$(-$ $(-$ $a$ $a$ $, 0)$ $,0)$ becomes $(-$ $(-$ $2$ $2$ $, 0)$ $,0)$

Next, draw the parabola. Remember that the parabola’s concavity should be opposite the directrix

Since the parabola is concave left, use $y^{2} = - 4$ $y^2=-4$ $a$ $a$ $x$ $x$

Finally, form the equation by substituting $a = 2$ $a=2$ to the chosen standard form

$y^{2}$ $y^2$ $=$ $=$ $- 4$ $-4$ $a$ $a$ $x$ $x$ Standard Form

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

$y^{2}$ $y^2$ $=$ $=$ $- 8 x$ $-8x$

$y^{2} = - 8 x$ $y^2=-8x$

Graph Parabolas Given the Vertex, Focus and Directrix 1

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Quiz summary

Information

1. Question

Hint

Correct!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

2. Question

Hint

Great Work!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

3. Question

Hint

Excellent!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

4. Question

Hint

Correct!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

Quizzes

$y$ $y$	$=$ $=$	$-$ $-$ $a$ $a$	Directrix Formula
$y$ $y$	$=$ $=$	$-$ $-$ $3$ $3$

$x^{2}$ $x^2$	$=$ $=$	$4$ $4$ $a$ $a$ $y$ $y$	Standard Form
$x^{2}$ $x^2$	$=$ $=$	$4$ $4$ $(3)$ $(3)$ $y$ $y$	Substitute $a = 3$ $a=3$
$x^{2}$ $x^2$	$=$ $=$	$12 y$ $12y$

$y$ $y$	$=$ $=$	$a$ $a$	Directrix Formula
$y$ $y$	$=$ $=$	$1$ $1$

$x^{2}$ $x^2$	$=$ $=$	$- 4$ $-4$ $a$ $a$ $y$ $y$	Standard Form
$x^{2}$ $x^2$	$=$ $=$	$- 4$ $-4$ $(1)$ $(1)$ $y$ $y$	Substitute $a = 1$ $a=1$
$x^{2}$ $x^2$	$=$ $=$	$- 4 y$ $-4y$

$y^{2}$ $y^2$	$=$ $=$	$4$ $4$ $a$ $a$ $x$ $x$	Standard Form
$y^{2}$ $y^2$	$=$ $=$	$4$ $4$ $(\frac{1}{4})$ $(1/4)$ $x$ $x$	Substitute $a = \frac{1}{4}$ $a=1/4$
$y^{2}$ $y^2$	$=$ $=$	$x$ $x$

$y^{2}$ $y^2$	$=$ $=$	$- 4$ $-4$ $a$ $a$ $x$ $x$	Standard Form
$y^{2}$ $y^2$	$=$ $=$	$- 4$ $-4$ $(2)$ $(2)$ $x$ $x$	Substitute $a = 2$ $a=2$
$y^{2}$ $y^2$	$=$ $=$	$- 8 x$ $-8x$