Graph Parabolas Given the Vertex, Focus and Directrix 2

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Question 1 of 4

1. Question

Write the equation of the parabola given the following:

Vertex `(-2,0)`

Directrix `y=4`

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

Hint

Help Video

Correct

Nice Job!

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`(-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

2. Question

Write the equation of the parabola given the following:

Vertex `(2,5)`

Directrix `x=5`

`(y-5)^2=-12(x-2)`
`(x-2)^2=4y`
`(y-2)^2=-12(x-5)`
`(x-2)^2=-3(y+3)`

Hint

Help Video

Correct

Well Done!

Incorrect

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

3. Question

Write the equation of the parabola given the following:

Focus `(2,-1)`

Directrix `x=-2`

`(y+1)^2=8x`
`(y+1)^2=4x`
`(y-1)^2=(x+1)`
`(y+1)^2=4(x-2)`

Hint

Help Video

Correct

Great Work!

Incorrect

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

4. Question

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)

Hint

Help Video

Correct

Keep Going!

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`

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