10.3 The parabola  (Page 3/11)

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Graph $\text{\hspace{0.17em}}{y}^{2}=-16x.\text{\hspace{0.17em}}$ Identify and label the focus, directrix, and endpoints of the latus rectum.

Focus: $\text{\hspace{0.17em}}\left(-4,0\right);\text{\hspace{0.17em}}$ Directrix: $\text{\hspace{0.17em}}x=4;\text{\hspace{0.17em}}$ Endpoints of the latus rectum: $\text{\hspace{0.17em}}\left(-4,±8\right)$

Graphing a parabola with vertex (0, 0) and the y -axis as the axis of symmetry

Graph $\text{\hspace{0.17em}}{x}^{2}=-6y.\text{\hspace{0.17em}}$ Identify and label the focus , directrix    , and endpoints of the latus rectum    .

The standard form that applies to the given equation is $\text{\hspace{0.17em}}{x}^{2}=4py.\text{\hspace{0.17em}}$ Thus, the axis of symmetry is the y -axis. It follows that:

• $-6=4p,$ so $\text{\hspace{0.17em}}p=-\frac{3}{2}.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}p<0,$ the parabola opens down.
• the coordinates of the focus are $\text{\hspace{0.17em}}\left(0,p\right)=\left(0,-\frac{3}{2}\right)$
• the equation of the directrix is $\text{\hspace{0.17em}}y=-p=\frac{3}{2}$
• the endpoints of the latus rectum can be found by substituting into the original equation, $\text{\hspace{0.17em}}\left(±3,-\frac{3}{2}\right)$

Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola    .

Graph $\text{\hspace{0.17em}}{x}^{2}=8y.\text{\hspace{0.17em}}$ Identify and label the focus, directrix, and endpoints of the latus rectum.

Focus: $\text{\hspace{0.17em}}\left(0,2\right);\text{\hspace{0.17em}}$ Directrix: $\text{\hspace{0.17em}}y=-2;\text{\hspace{0.17em}}$ Endpoints of the latus rectum: $\text{\hspace{0.17em}}\left(±4,2\right).$

Writing 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.

Given its focus and directrix, write the equation for a parabola in standard form.

1. Determine whether the axis of symmetry is the x - or y -axis.
1. If the given coordinates of the focus have the form $\text{\hspace{0.17em}}\left(p,0\right),$ then the axis of symmetry is the x -axis. Use the standard form $\text{\hspace{0.17em}}{y}^{2}=4px.$
2. If the given coordinates of the focus have the form $\text{\hspace{0.17em}}\left(0,p\right),$ then the axis of symmetry is the y -axis. Use the standard form $\text{\hspace{0.17em}}{x}^{2}=4py.$
2. Multiply $\text{\hspace{0.17em}}4p.$
3. Substitute the value from Step 2 into the equation determined in Step 1.

Writing the equation of a parabola in standard form given its focus and directrix

What is the equation for the parabola    with focus $\text{\hspace{0.17em}}\left(-\frac{1}{2},0\right)\text{\hspace{0.17em}}$ and directrix     $\text{\hspace{0.17em}}x=\frac{1}{2}?$

The focus has the form $\text{\hspace{0.17em}}\left(p,0\right),$ so the equation will have the form $\text{\hspace{0.17em}}{y}^{2}=4px.$

• Multiplying $\text{\hspace{0.17em}}4p,$ we have $\text{\hspace{0.17em}}4p=4\left(-\frac{1}{2}\right)=-2.$
• Substituting for $\text{\hspace{0.17em}}4p,$ we have $\text{\hspace{0.17em}}{y}^{2}=4px=-2x.$

Therefore, the equation for the parabola is $\text{\hspace{0.17em}}{y}^{2}=-2x.$

What is the equation for the parabola with focus $\text{\hspace{0.17em}}\left(0,\frac{7}{2}\right)\text{\hspace{0.17em}}$ and directrix $\text{\hspace{0.17em}}y=-\frac{7}{2}?$

${x}^{2}=14y.$

Graphing parabolas with vertices not at the origin

Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ units horizontally and $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ units vertically, the vertex will be $\text{\hspace{0.17em}}\left(h,k\right).\text{\hspace{0.17em}}$ This translation results in the standard form of the equation we saw previously with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ replaced by $\text{\hspace{0.17em}}\left(x-h\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ replaced by $\text{\hspace{0.17em}}\left(y-k\right).$

To graph parabolas with a vertex $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ other than the origin, we use the standard form $\text{\hspace{0.17em}}{\left(y-k\right)}^{2}=4p\left(x-h\right)\text{\hspace{0.17em}}$ for parabolas that have an axis of symmetry parallel to the x -axis, and $\text{\hspace{0.17em}}{\left(x-h\right)}^{2}=4p\left(y-k\right)\text{\hspace{0.17em}}$ for parabolas that have an axis of symmetry parallel to the y -axis. These standard forms are given below, along with their general graphs and key features.

Standard forms of parabolas with vertex ( h , k )

[link] and [link] summarize the standard features of parabolas with a vertex at a point $\text{\hspace{0.17em}}\left(h,k\right).$

 Axis of Symmetry Equation Focus Directrix Endpoints of Latus Rectum $y=k$ ${\left(y-k\right)}^{2}=4p\left(x-h\right)$ $x=h-p$ $x=h$ ${\left(x-h\right)}^{2}=4p\left(y-k\right)$ $y=k-p$

how can are find the domain and range of a relations
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6000
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Robert
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with doing calculus
SLIMANE
Thanks po.
Jenica
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Marco
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Jenica
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SLIMANE
What is domain
johnphilip
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Chris
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Jeffrey
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meena
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meena
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I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
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