# 12.3 The parabola  (Page 6/11)

 Page 6 / 11

## Key equations

 Parabola, vertex at origin, axis of symmetry on x -axis ${y}^{2}=4px$ Parabola, vertex at origin, axis of symmetry on y -axis ${x}^{2}=4py$ Parabola, vertex at $\text{\hspace{0.17em}}\left(h,k\right),$ axis of symmetry on x -axis ${\left(y-k\right)}^{2}=4p\left(x-h\right)$ Parabola, vertex at $\text{\hspace{0.17em}}\left(h,k\right),$ axis of symmetry on y -axis ${\left(x-h\right)}^{2}=4p\left(y-k\right)$

## Key concepts

• A parabola is the set of all points $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.
• The standard form of a parabola with vertex $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and the x -axis as its axis of symmetry can be used to graph the parabola. If $\text{\hspace{0.17em}}p>0,$ the parabola opens right. If $\text{\hspace{0.17em}}p<0,$ the parabola opens left. See [link] .
• The standard form of a parabola with vertex $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and the y -axis as its axis of symmetry can be used to graph the parabola. If $\text{\hspace{0.17em}}p>0,$ the parabola opens up. If $\text{\hspace{0.17em}}p<0,$ the parabola opens down. See [link] .
• When given the focus and directrix of a parabola, we can write its equation in standard form. See [link] .
• The standard form of a parabola with vertex $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and axis of symmetry parallel to the x -axis can be used to graph the parabola. If $\text{\hspace{0.17em}}p>0,$ the parabola opens right. If $\text{\hspace{0.17em}}p<0,$ the parabola opens left. See [link] .
• The standard form of a parabola with vertex $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and axis of symmetry parallel to the y -axis can be used to graph the parabola. If $\text{\hspace{0.17em}}p>0,$ the parabola opens up. If $\text{\hspace{0.17em}}p<0,$ the parabola opens down. See [link] .
• Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides. See [link] .

## Verbal

Define a parabola in terms of its focus and directrix.

A parabola is the set of points in the plane that lie equidistant from a fixed point, the focus, and a fixed line, the directrix.

If the equation of a parabola is written in standard form and $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is positive and the directrix is a vertical line, then what can we conclude about its graph?

If the equation of a parabola is written in standard form and $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is negative and the directrix is a horizontal line, then what can we conclude about its graph?

The graph will open down.

What is the effect on the graph of a parabola if its equation in standard form has increasing values of $\text{\hspace{0.17em}}p\text{?}$

As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?

The distance between the focus and directrix will increase.

## Algebraic

For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form.

${y}^{2}=4-{x}^{2}$

$y=4{x}^{2}$

yes $\text{\hspace{0.17em}}y=4\left(1\right){x}^{2}$

$3{x}^{2}-6{y}^{2}=12$

${\left(y-3\right)}^{2}=8\left(x-2\right)$

yes $\text{\hspace{0.17em}}{\left(y-3\right)}^{2}=4\left(2\right)\left(x-2\right)$

${y}^{2}+12x-6y-51=0$

For the following exercises, rewrite the given equation in standard form, and then determine the vertex $\text{\hspace{0.17em}}\left(V\right),$ focus $\text{\hspace{0.17em}}\left(F\right),$ and directrix of the parabola.

$x=8{y}^{2}$

${y}^{2}=\frac{1}{8}x,V:\left(0,0\right);F:\left(\frac{1}{32},0\right);d:x=-\frac{1}{32}$

$y=\frac{1}{4}{x}^{2}$

$y=-4{x}^{2}$

${x}^{2}=-\frac{1}{4}y,V:\left(0,0\right);F:\left(0,-\frac{1}{16}\right);d:y=\frac{1}{16}$

$x=\frac{1}{8}{y}^{2}$

$x=36{y}^{2}$

${y}^{2}=\frac{1}{36}x,V:\left(0,0\right);F:\left(\frac{1}{144},0\right);d:x=-\frac{1}{144}$

$x=\frac{1}{36}{y}^{2}$

${\left(x-1\right)}^{2}=4\left(y-1\right)$

${\left(x-1\right)}^{2}=4\left(y-1\right),V:\left(1,1\right);F:\left(1,2\right);d:y=0$

${\left(y-2\right)}^{2}=\frac{4}{5}\left(x+4\right)$

${\left(y-4\right)}^{2}=2\left(x+3\right)$

${\left(y-4\right)}^{2}=2\left(x+3\right),V:\left(-3,4\right);F:\left(-\frac{5}{2},4\right);d:x=-\frac{7}{2}$

${\left(x+1\right)}^{2}=2\left(y+4\right)$

${\left(x+4\right)}^{2}=24\left(y+1\right)$

${\left(x+4\right)}^{2}=24\left(y+1\right),V:\left(-4,-1\right);F:\left(-4,5\right);d:y=-7$

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