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Graph Parabolas Given the Vertex, Focus and Directrix>
Graph Parabolas Given the Vertex, Focus and Directrix 1Graph 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)`Vertex `(0,0)`- `x^2=` (12)`y`
Hint
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Correct!
Incorrect
Standard Form (Concave Up)
`x^2=4``a``y`Focus
`(0,``a``)`Standard Form (Concave Down)
`x^2=-4``a``y`Focus
`(0,-``a``)`First, plot the given focus and vertexFocus`(0,3)`Vertex`(0,0)`
Based on the diagram, the focal length, `a`, is equal to `3`Now that we know `a`, we can also plot the directrix.`y` `=` `-``a` Directrix Formula `y` `=` `-``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``a``y`Finally, form the equation by substituting `a=3` to the chosen standard form`x^2` `=` `4``a``y` Standard Form `x^2` `=` `4``(3)``y` Substitute `a=3` `x^2` `=` `12y` `x^2=12y` -
Question 2 of 4
2. Question
Write the equation of the parabola given the following:Focus `(0,-1)`Vertex `(0,0)`- `x^2=` (-4)`y`
Hint
Help VideoCorrect
Great Work!
Incorrect
Standard Form (Concave Up)
`x^2=4``a``y`Focus
`(0,``a``)`Standard Form (Concave Down)
`x^2=-4``a``y`Focus
`(0,-``a``)`First, plot the given focus and vertexFocus`(0,-1)`Vertex`(0,0)`
Based on the diagram, the focal length, `a`, is equal to `1`Now that we know `a`, we can also plot the directrix.`y` `=` `a` Directrix Formula `y` `=` `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``a``y`Finally, form the equation by substituting `a=1` to the chosen standard form`x^2` `=` `-4``a``y` Standard Form `x^2` `=` `-4``(1)``y` Substitute `a=1` `x^2` `=` `-4y` `x^2=-4y` -
Question 3 of 4
3. Question
Write the equation of the parabola given the following:Directrix `x=-1/4`Vertex `(0,0)`- `y^2=` (x)
Hint
Help VideoCorrect
Excellent!
Incorrect
Standard Form (Concave Right)
`y^2=4``a``x`Focus
`(``a``,0)`Standard Form (Concave Left)
`y^2=-4``a``x`Focus
`(-``a``,0)`First, plot the given directrix and vertexDirectrix `x=-1/4`Vertex`(0,0)`
Based on the diagram, the focal length, `a`, is equal to `1/4`Now that we know `a`, we can also plot the focus.`(``a``,0)` becomes `(``1/4``,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``a``x`Finally, form the equation by substituting `a=1/4` to the chosen standard form`y^2` `=` `4``a``x` Standard Form `y^2` `=` `4``(1/4)``x` Substitute `a=1/4` `y^2` `=` `x` `y^2=x` -
Question 4 of 4
4. Question
Write the equation of the parabola given the following:Directrix `x=2`Vertex `(0,0)`- `y^2=` (-8)`x`
Hint
Help VideoCorrect
Correct!
Incorrect
Standard Form (Concave Right)
`y^2=4``a``x`Focus
`(``a``,0)`Standard Form (Concave Left)
`y^2=-4``a``x`Focus
`(-``a``,0)`First, plot the given directrix and vertexDirectrix `x=2`Vertex`(0,0)`
Based on the diagram, the focal length, `a`, is equal to `2`Now that we know `a`, we can also plot the focus.`(-``a``,0)` becomes `(-``2``,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``a``x`Finally, form the equation by substituting `a=2` to the chosen standard form`y^2` `=` `-4``a``x` Standard Form `y^2` `=` `-4``(2)``x` Substitute `a=2` `y^2` `=` `-8x` `y^2=-8x`
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