Graph Parabolas Given the Vertex, Focus and Directrix 2

Question 1 of 4

Write the equation of the parabola given the following:

Vertex

$(-2,0)$

Directrix

$y=4$

1. $(x-2)^2=16y$
2. $(x+2)^2=16y$
3. $x^2=-16y$
4. $(x+2)^2=-16y$

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Standard Form (Concave Up)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,$ $a$

$+$ $k$

$)$

Standard Form (Concave Down)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,-$ $a$

$+$ $k$

$)$

First, plot the given vertex and directrix

Vertex

$(-2,0)$

Directrix

$y=4$

This also means that $h=-2$ and $k=0$

Remember that the perpendicular distance between the vertex and directrix is the focal length $(a)$

Find this distance by subtracting $k=0$ from the directrix

$(y=4)$ .

Distance	$=$	$4-0$
$a$	$=$	$4$

Now that we know

$a$ , we can also plot the focus which should be $4$ units below the vertex.

Focus	$=$	$(-2,0-$ $4$ $)$
	$=$	$(-2,-4)$

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

Since the parabola is concave down, use

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Finally, form the equation by substituting $a=4$ , $h=-2$ and $k=0$ to the chosen standard form

${(x-\color{#00880a}{h})}^2$	$=$	$-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$	Standard Form
${(x-\color{#00880a}{(-2)})}^2$	$=$	$-4\color{#9a00c7}{(4)}(y-\color{#007ddc}{0})$	Substitute $a=4$ , $h=-2$ and $k=0$
$(x+2)^2$	$=$	$-16y$

$(x+2)^2=-16y$

Question 2 of 4

Write the equation of the parabola given the following:

Vertex

$(2,5)$

Directrix

$x=5$

1. $(y-5)^2=-12(x-2)$
2. $(x-2)^2=4y$
3. $(x-2)^2=-3(y+3)$
4. $(y-2)^2=-12(x-5)$

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Standard Form (Concave Right)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$($ $a$

$+$ $h$

$,$ $k$

$)$

Standard Form (Concave Left)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$(-$ $a$

$+$ $h$

$,$ $k$

$)$

First, plot the given vertex and directrix

Vertex

$(2,5)$

Directrix

$x=5$

This also means that $h=2$ and $k=5$

Remember that the perpendicular distance between the vertex and directrix is the focal length $(a)$

Find this distance by subtracting $h=2$ from the directrix

$(x=5)$ .

Distance	$=$	$5-2$
$a$	$=$	$3$

Now that we know

$a$ , we can also plot the focus which should be $3$ units to the left the vertex.

Focus	$=$	$(2-$ $3$ $,5)$
	$=$	$(-1,5)$

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

Since the parabola is concave left, use

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Finally, form the equation by substituting $a=3$ , $h=2$ and $k=5$ to the chosen standard form

${(y-\color{#007ddc}{k})}^2$	$=$	$-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$	Standard Form
${(y-\color{#007ddc}{5})}^2$	$=$	$-4\color{#9a00c7}{(3)}(x-\color{#00880a}{2})$	Substitute $a=3$ , $h=2$ and $k=5$
$(y-5)^2$	$=$	$-12(x-2)$

$(y-5)^2=-12(x-2)$

Question 3 of 4

Write the equation of the parabola given the following:

Focus

$(2,-1)$

Directrix

$x=-2$

1. $(y+1)^2=8x$
2. $(y+1)^2=4x$
3. $(y-1)^2=(x+1)$
4. $(y+1)^2=4(x-2)$

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Standard Form (Concave Right)

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$($ $a$

$+$ $h$

$,$ $k$

$)$

Standard Form (Concave Left)

${(y-\color{#007ddc}{k})}^2=-4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Focus

$(-$ $a$

$+$ $h$

$,$ $k$

$)$

First, plot the given focus and directrix

Focus

$(2,-1)$

Directrix

$x=-2$

Remember that the perpendicular distance between the focus and directrix is twice the focal length

$(2$ $a$

$)$

Find this distance by subtracting the directrix from the focus’

$x$ value.

Distance	$=$	$2-(-2)$
	$=$	$4$

Equate this distance to

$2a$

$2a$	$=$	$4$
$2a$ $divide2$	$=$	$4$ $divide2$	Divide both sides by $2$
$a$	$=$	$2$

Now that we know

$a$ , we can also plot the vertex which is $2$ units to the left of the focus.

Vertex	$=$	$(2-$ $2$ $,-1)$
	$=$	$(0,-1)$

This means that $h=0$ and $k=-1$

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

Since the parabola is concave right, use

${(y-\color{#007ddc}{k})}^2=4\color{#9a00c7}{a}(x-\color{#00880a}{h})$

Finally, form the equation by substituting $a=2$ , $h=0$ and $k=-1$ to the chosen standard form

${(y-\color{#007ddc}{k})}^2$	$=$	$4\color{#9a00c7}{a}(x-\color{#00880a}{h})$	Standard Form
${(y-\color{#007ddc}{(-1)})}^2$	$=$	$4\color{#9a00c7}{(2)}(x-\color{#00880a}{0})$	Substitute $a=2$ , $h=0$ and $k=-1$
$(y+1)^2$	$=$	$8x$

$(y+1)^2=8x$

Question 4 of 4

Write the equation of the parabola given the following:

Vertex

$(-1,3)$

Directrix

$y=3.5$

Write fractions as “a/b”

$a=$ (-1/2)

$b=$ (-1)

$c=$ (5/2)

Incorrect

Standard Form (Concave Up)

${(x-\color{#00880a}{h})}^2=4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,$ $a$

$+$ $k$

$)$

Standard Form (Concave Down)

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Focus

$($ $h$

$,-$ $a$

$+$ $k$

$)$

First, plot the given vertex and directrix

Vertex

$(-1,3)$

Directrix

$y=3.5$ or

$y=3 1/2$

This also means that $h=-1$ and $k=3$

Remember that the perpendicular distance between the vertex and directrix is the focal length $(a)$

Find this distance by subtracting $k=3$ from the directrix

$(y=3 1/2)$ .

Distance	$=$	$3 1/2-3$

$a$	$=$	$1/2$

Now that we know

$a$ , we can also plot the focus which should be $1/2$ unit below the vertex.

Focus	$=$	$(-1,3-$ $1/2$ $)$

	$=$	$(-1,2 1/2)$

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

Since the parabola is concave down, use

${(x-\color{#00880a}{h})}^2=-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$

Finally, form the equation by substituting $a=1/2$ , $h=-1$ and $k=3$ to the chosen standard form

${(x-\color{#00880a}{h})}^2$	$=$	$-4\color{#9a00c7}{a}(y-\color{#007ddc}{k})$	Standard Form

${(x-\color{#00880a}{(-1)})}^2$	$=$	$-4\color{#9a00c7}{\frac{1}{2}}(y-\color{#007ddc}{3})$	Substitute $a=1/2$ , $h=-1$ and $k=3$

$(x+1)^2$	$=$	$-2(y-3)$

Arrange the equation in the form

$y=ax^2+bx+c$ to get the value of

$a$ ,

$b$ and

$c$

$(x+1)^2$	$=$	$-2(y-3)$
$x^2+2x+1$	$=$	$-2y+6$	Expand
$-2y+6$	$=$	$x^2+2x+1$	Move $y$ term to the left
$-2y+6$ $-6$	$=$	$x^2+2x+1$ $-6$	Subtract $6$ from both sides
$-2y$	$=$	$x^2+2x-5$
$-2y$ $divide-2$	$=$	$(x^2+2x-5)$ $divide-2$	Divide both sides by $-2$

$y$	$=$	$-(x^2)/2-x+5/2$

Therefore,

$a=-1/2$ ,

$b=-1$ and

$c=5/2$

$a=-1/2$

$b=-1$

$c=5/2$

Graph Parabolas Given the Vertex, Focus and Directrix 2

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Quiz summary

Information

1. Question

Hint

Nice Job!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

2. Question

Hint

Well Done!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

3. Question

Hint

Great Work!

Standard Form (Concave Right)

Focus

Standard Form (Concave Left)

Focus

4. Question

Hint

Keep Going!

Standard Form (Concave Up)

Focus

Standard Form (Concave Down)

Focus

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