# 12.3 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 $y=k$ ${\left(y-k\right)}^{2}=4p\left(x-h\right)$ $x=h-p$ $x=h$ ${\left(x-h\right)}^{2}=4p\left(y-k\right)$ $y=k-p$

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
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