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Locus and The Parabola>
Graphing Parabolas Given the Vertex, Focus and Directrix>
Graphing Parabolas Given the Vertex, Focus and Directrix 2Graphing 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`Hint
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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 directrixVertex`(-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`Hint
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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 directrixVertex`(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`Hint
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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 directrixFocus`(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
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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 directrixVertex`(-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` -