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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 $\text{\hspace{0.17em}}(h,k),$ axis of symmetry on x -axis | ${\left(y-k\right)}^{2}=4p\left(x-h\right)$ |
Parabola, vertex at $\text{\hspace{0.17em}}(h,k),$ axis of symmetry on y -axis | ${\left(x-h\right)}^{2}=4p\left(y-k\right)$ |
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.
For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form.
${y}^{2}=4-{x}^{2}$
$3{x}^{2}-6{y}^{2}=12$
${\left(y-3\right)}^{2}=8\left(x-2\right)$
yes $\text{\hspace{0.17em}}{\left(y-3\right)}^{2}=4(2)\left(x-2\right)$
${y}^{2}+12x-6y-51=0$
For the following exercises, rewrite the given equation in standard form, and then determine the vertex $\text{\hspace{0.17em}}(V),$ focus $\text{\hspace{0.17em}}(F),$ and directrix $\text{}(d)\text{}$ of the parabola.
$x=8{y}^{2}$
${y}^{2}=\frac{1}{8}x,V:(0,0);F:\left(\frac{1}{32},0\right);d:x=-\frac{1}{32}$
$y=\frac{1}{4}{x}^{2}$
$y=\mathrm{-4}{x}^{2}$
${x}^{2}=-\frac{1}{4}y,V:\left(0,0\right);F:\left(0,-\frac{1}{16}\right);d:y=\frac{1}{16}$
$x=\frac{1}{8}{y}^{2}$
$x=36{y}^{2}$
${y}^{2}=\frac{1}{36}x,V:\left(0,0\right);F:\left(\frac{1}{144},0\right);d:x=-\frac{1}{144}$
$x=\frac{1}{36}{y}^{2}$
${\left(x-1\right)}^{2}=4\left(y-1\right)$
${\left(x-1\right)}^{2}=4\left(y-1\right),V:\left(1,1\right);F:\left(1,2\right);d:y=0$
${\left(y-2\right)}^{2}=\frac{4}{5}\left(x+4\right)$
${\left(y-4\right)}^{2}=2\left(x+3\right)$
${\left(y-4\right)}^{2}=2\left(x+3\right),V:\left(-3,4\right);F:\left(-\frac{5}{2},4\right);d:x=-\frac{7}{2}$
${\left(x+1\right)}^{2}=2\left(y+4\right)$
${\left(x+4\right)}^{2}=24\left(y+1\right)$
${\left(x+4\right)}^{2}=24\left(y+1\right),V:\left(-4,-1\right);F:\left(-4,5\right);d:y=\mathrm{-7}$
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