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For the following exercises, determine the equation of the parabola using the information given.
Focus $\left(4,0\right)$ and directrix $x=\mathrm{-4}$
${y}^{2}=16x$
Focus $\left(0,\mathrm{-3}\right)$ and directrix $y=3$
Focus $\left(0,0.5\right)$ and directrix $y=\mathrm{-0.5}$
${x}^{2}=2y$
Focus $\left(2,\phantom{\rule{0.2em}{0ex}}3\right)$ and directrix $x=\mathrm{-2}$
Focus $\left(0,2\right)$ and directrix $y=4$
${x}^{2}=\mathrm{-4}\left(y-3\right)$
Focus $\left(\mathrm{-1},4\right)$ and directrix $x=5$
Focus $\left(\mathrm{-3},5\right)$ and directrix $y=1$
${\left(x+3\right)}^{2}=8\left(y-3\right)$
Focus $\left(\frac{5}{2},\mathrm{-4}\right)$ and directrix $x=\frac{7}{2}$
For the following exercises, determine the equation of the ellipse using the information given.
Endpoints of major axis at $\left(4,0\right),\left(\mathrm{-4},0\right)$ and foci located at $\left(2,0\right),\left(\mathrm{-2},0\right)$
$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{12}=1$
Endpoints of major axis at $\left(0,5\right),\left(0,\mathrm{-5}\right)$ and foci located at $\left(0,3\right),\left(0,\mathrm{-3}\right)$
Endpoints of major axis at $\left(0,2\right),\left(0,\mathrm{-2}\right)$ and foci located at $\left(3,0\right),\left(\mathrm{-3},0\right)$
$\frac{{x}^{2}}{13}+\frac{{y}^{2}}{4}=1$
Endpoints of major axis at $\left(\mathrm{-3},3\right),\left(7,3\right)$ and foci located at $\left(\mathrm{-2},3\right),\left(6,3\right)$
Endpoints of major axis at $\left(\mathrm{-3},5\right),\left(\mathrm{-3},\mathrm{-3}\right)$ and foci located at $\left(\mathrm{-3},3\right),\left(\mathrm{-3},\mathrm{-1}\right)$
$\frac{{\left(y-1\right)}^{2}}{16}+\frac{{\left(x+3\right)}^{2}}{12}=1$
Endpoints of major axis at $\left(0,0\right),\left(0,4\right)$ and foci located at $\left(5,2\right),\left(\mathrm{-5},2\right)$
Foci located at $\left(2,0\right),\phantom{\rule{0.2em}{0ex}}\left(\mathrm{-2},0\right)$ and eccentricity of $\frac{1}{2}$
$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{12}=1$
Foci located at $\left(0,\mathrm{-3}\right),\phantom{\rule{0.2em}{0ex}}\left(0,3\right)$ and eccentricity of $\frac{3}{4}$
For the following exercises, determine the equation of the hyperbola using the information given.
Vertices located at $\left(5,0\right),\left(\mathrm{-5},0\right)$ and foci located at $\left(6,0\right),\left(\mathrm{-6},0\right)$
$\frac{{x}^{2}}{25}-\frac{{y}^{2}}{11}=1$
Vertices located at $\left(0,2\right),\left(0,\mathrm{-2}\right)$ and foci located at $\left(0,3\right),\left(0,\mathrm{-3}\right)$
Endpoints of the conjugate axis located at $\left(0,3\right),\left(0,\mathrm{-3}\right)$ and foci located $\left(4,0\right),\left(\mathrm{-4},0\right)$
$\frac{{x}^{2}}{7}-\frac{{y}^{2}}{9}=1$
Vertices located at $\left(0,1\right),\left(6,1\right)$ and focus located at $\left(8,1\right)$
Vertices located at $\left(\mathrm{-2},0\right),\left(\mathrm{-2},\mathrm{-4}\right)$ and focus located at $\left(\mathrm{-2},\mathrm{-8}\right)$
$\frac{{\left(y+2\right)}^{2}}{4}-\frac{{\left(x+2\right)}^{2}}{32}=1$
Endpoints of the conjugate axis located at $\left(3,2\right),\left(3,4\right)$ and focus located at $\left(3,7\right)$
Foci located at $\left(6,\mathrm{-0}\right),\left(6,0\right)$ and eccentricity of 3
$\frac{{x}^{2}}{4}-\frac{{y}^{2}}{32}=1$
$\left(0,10\right),\left(0,\mathrm{-10}\right)$ and eccentricity of 2.5
For the following exercises, consider the following polar equations of conics. Determine the eccentricity and identify the conic.
$r=\frac{\mathrm{-1}}{1+\text{cos}\phantom{\rule{0.2em}{0ex}}\theta}$
$e=1,$ parabola
$r=\frac{8}{2-\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{5}{2+\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$e=\frac{1}{2},$ ellipse
$r=\frac{5}{\mathrm{-1}+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{3}{2-6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$e=3,$ hyperbola
$r=\frac{3}{\mathrm{-4}+3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.
$\text{Directrix:}\phantom{\rule{0.2em}{0ex}}x=4;e=\frac{1}{5}$
$r=\frac{4}{5+\text{cos}\phantom{\rule{0.2em}{0ex}}\theta}$
$\text{Directrix:}\phantom{\rule{0.2em}{0ex}}x=\mathrm{-4};e=5$
$\text{Directrix: y}=2;e=2$
$r=\frac{4}{1+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$\text{Directrix: y}=\mathrm{-2};e=\frac{1}{2}$
For the following exercises, sketch the graph of each conic.
$r=\frac{1}{1-\text{cos}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{10}{5+4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{15}{3-2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{32}{3+5\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{3}{2+6\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$r=\frac{3}{\mathrm{-4}+2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta}$
$\begin{array}{c}\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1\hfill \\ \hfill \end{array}$
$4{x}^{2}+9{y}^{2}=36$
$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{9}=1$
${y}^{2}=20x$
For the following equations, determine which of the conic sections is described.
${x}^{2}+2\sqrt{3}xy+3{y}^{2}-6=0$
$34{x}^{2}-24xy+41{y}^{2}-25=0$
The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given as ${x}^{2}=4y.$ At what coordinates should you place the lightbulb?
A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?
At the point 2.25 feet above the vertex.
Consider the satellite dish of the preceding problem. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?
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