Graphing Parabolas with Vertices at the Origin
In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola.
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[latex]\begin{array}{l}d=\sqrt{{\left(x - 0\right)}^{2}+{\left(y-p\right)}^{2}}\hfill \\ =\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}\hfill \end{array}[/latex]
Set the two expressions for [latex]d[/latex] equal to each other and solve for [latex]y[/latex] to derive the equation of the parabola. We do this because the distance from [latex]\left(x,y\right)[/latex] to [latex]\left(0,p\right)[/latex] equals the distance from [latex]\left(x,y\right)[/latex] to [latex]\left(x, -p\right)[/latex].
[latex]\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}=y+p[/latex]
We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.
[latex]\begin{array}{c}{x}^{2}+{\left(y-p\right)}^{2}={\left(y+p\right)}^{2}\\ {x}^{2}+{y}^{2}-2py+{p}^{2}={y}^{2}+2py+{p}^{2}\\ {x}^{2}-2py=2py\\ \text{ }{x}^{2}=4py\end{array}[/latex]
The equations of parabolas with vertex [latex]\left(0,0\right)[/latex] are [latex]{y}^{2}=4px[/latex] when the x-axis is the axis of symmetry and [latex]{x}^{2}=4py[/latex] when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
A General Note: Standard Forms of Parabolas with Vertex (0, 0)
The table below and Figure 5 summarize the standard features of parabolas with a vertex at the origin.Axis of Symmetry | Equation | Focus | Directrix | Endpoints of Latus Rectum |
x-axis | [latex]{y}^{2}=4px[/latex] | [latex]\left(p,\text{ }0\right)[/latex] | [latex]x=-p[/latex] | [latex]\left(p,\text{ }\pm 2p\right)[/latex] |
y-axis | [latex]{x}^{2}=4py[/latex] | [latex]\left(0,\text{ }p\right)[/latex] | [latex]y=-p[/latex] | [latex]\left(\pm 2p,\text{ }p\right)[/latex] |
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How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.
- Determine which of the standard forms applies to the given equation: [latex]{y}^{2}=4px[/latex] or [latex]{x}^{2}=4py[/latex].
- Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
- If the equation is in the form [latex]{y}^{2}=4px[/latex], then
- the axis of symmetry is the x-axis, [latex]y=0[/latex]
- set [latex]4p[/latex] equal to the coefficient of x in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens right. If [latex]p<0[/latex], the parabola opens left.
- use [latex]p[/latex] to find the coordinates of the focus, [latex]\left(p,0\right)[/latex]
- use [latex]p[/latex] to find the equation of the directrix, [latex]x=-p[/latex]
- use [latex]p[/latex] to find the endpoints of the latus rectum, [latex]\left(p,\pm 2p\right)[/latex]. Alternately, substitute [latex]x=p[/latex] into the original equation.
- If the equation is in the form [latex]{x}^{2}=4py[/latex], then
- the axis of symmetry is the y-axis, [latex]x=0[/latex]
- set [latex]4p[/latex] equal to the coefficient of y in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens up. If [latex]p<0[/latex], the parabola opens down.
- use [latex]p[/latex] to find the coordinates of the focus, [latex]\left(0,p\right)[/latex]
- use [latex]p[/latex] to find equation of the directrix, [latex]y=-p[/latex]
- use [latex]p[/latex] to find the endpoints of the latus rectum, [latex]\left(\pm 2p,p\right)[/latex]
- If the equation is in the form [latex]{y}^{2}=4px[/latex], then
- Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Example 1: Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry
Graph [latex]{y}^{2}=24x[/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.Solution
The standard form that applies to the given equation is [latex]{y}^{2}=4px[/latex]. Thus, the axis of symmetry is the x-axis. It follows that:- [latex]24=4p[/latex], so [latex]p=6[/latex]. Since [latex]p>0[/latex], the parabola opens right the coordinates of the focus are [latex]\left(p,0\right)=\left(6,0\right)[/latex]
- the equation of the directrix is [latex]x=-p=-6[/latex]
- the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute [latex]x=6[/latex] into the original equation: [latex]\left(6,\pm 12\right)[/latex]
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Try It 1
Graph [latex]{y}^{2}=-16x[/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum. SolutionExample 2: Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry
Graph [latex]{x}^{2}=-6y[/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum.Solution
The standard form that applies to the given equation is [latex]{x}^{2}=4py[/latex]. Thus, the axis of symmetry is the y-axis. It follows that:- [latex]-6=4p[/latex], so [latex]p=-\frac{3}{2}[/latex]. Since [latex]p<0[/latex], the parabola opens down.
- the coordinates of the focus are [latex]\left(0,p\right)=\left(0,-\frac{3}{2}\right)[/latex]
- the equation of the directrix is [latex]y=-p=\frac{3}{2}[/latex]
- the endpoints of the latus rectum can be found by substituting [latex]\text{ }y=\frac{3}{2}\text{ }[/latex] into the original equation, [latex]\left(\pm 3,-\frac{3}{2}\right)[/latex]
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Try It 2
Graph [latex]{x}^{2}=8y[/latex]. Identify and label the focus, directrix, and endpoints of the latus rectum. SolutionWriting Equations of Parabolas in Standard Form
In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.How To: Given its focus and directrix, write the equation for a parabola in standard form.
- Determine whether the axis of symmetry is the x- or y-axis.
- If the given coordinates of the focus have the form [latex]\left(p,0\right)[/latex], then the axis of symmetry is the x-axis. Use the standard form [latex]{y}^{2}=4px[/latex].
- If the given coordinates of the focus have the form [latex]\left(0,p\right)[/latex], then the axis of symmetry is the y-axis. Use the standard form [latex]{x}^{2}=4py[/latex].
- Multiply [latex]4p[/latex].
- Substitute the value from Step 2 into the equation determined in Step 1.
Example 3: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix
What is the equation for the parabola with focus [latex]\left(-\frac{1}{2},0\right)[/latex] and directrix [latex]x=\frac{1}{2}?[/latex]Solution
The focus has the form [latex]\left(p,0\right)[/latex], so the equation will have the form [latex]{y}^{2}=4px[/latex].Multiplying [latex]4p[/latex], we have [latex]4p=4\left(-\frac{1}{2}\right)=-2[/latex]. Substituting for [latex]4p[/latex], we have [latex]{y}^{2}=4px=-2x[/latex].