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# 7.E: Hooked on Conics (Exercises)

[ "article:topic", "authorname:stitzzeager" ]

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## 7.1: Introduction to Conics

In Exercises \ref{circleeqnfirst} - \ref{circleeqnlast}, find the standard equation of the circle and then graph it.

\begin{multicols}{2}

\begin{enumerate}

\item Center $(-1, -5)$, radius $10$ \label{circleeqnfirst}

\item Center $(4,-2)$, radius $3$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Center $\left(-3, \frac{7}{13}\right)$, radius $\frac{1}{2}$

\item Center $(5, -9)$, radius $\ln(8)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Center $\left(-e, \sqrt{2}\right)$, radius $\pi$

\item Center $(\pi, e^{2})$, radius $\sqrt[3]{91}$ \label{circleeqnlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{ctscirclefirst} - \ref{ctscirclelast}, complete the square in order to put the equation into standard form. Identify the center and the radius or explain why the equation does not represent a circle.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^{2} - 4x + y^{2} + 10y = -25$ \label{ctscirclefirst}

\item $-2x^{2} - 36x - 2y^{2} - 112 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2+y^2+8x-10y -1 =0$

\item $x^2+y^2+5x-y-1=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $4x^{2} + 4y^{2} - 24y + 36 = 0$

\item $x^{2} + x + y^{2} - \frac{6}{5}y = 1$ \label{ctscirclelast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

## 7.2: Circles

In Exercises \ref{buildcirclefirst} - \ref{buildcirclelast}, find the standard equation of the circle which satisfies the given criteria.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

1. \item center $(3, 5)$, passes through $(-1, -2)$ \label{buildcirclefirst}
2. \item center $(3, 6)$, passes through $(-1, 4)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item endpoints of a diameter: $(3,6)$ and $(-1,4)$

\item endpoints of a diameter: $\left( \frac{1}{2}, 4\right)$, $\left(\frac{3}{2}, -1\right)$ \label{buildcirclelast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height is 136 feet.\footnote{Source: \href{http://www.cedarpoint.com/public/par...nderline{Cedar Point's webpage}}.} Find an equation for the wheel assuming that its center lies on the $y$-axis.

\label{giantwheelcircle}

1. \item Verify that the following points lie on the Unit Circle: $(\pm 1, 0)$, $(0, \pm 1)$, $\left(\pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}\right)$, $\left(\pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)$ and $\left(\pm \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\right)$
2. \item \label{circletransunitcircleexercise} Discuss with your classmates how to obtain the standard equation of a circle, Equation \ref{standardcircle}, from the equation of the Unit Circle, $x^2+y^2=1$ using the transformations discussed in Section \ref{Transformations}. (Thus every circle is just a few transformations away from the Unit Circle.)
3. \item Find an equation for the function represented graphically by the top half of the Unit Circle. Explain how the transformations is Section \ref{Transformations} can be used to produce a function whose graph is either the top or bottom of an arbitrary circle.
4. \item Find a one-to-one function whose graph is half of a circle. (Hint: Think piecewise.)

\begin{multicols}{2}

\begin{enumerate}

\item $(x + 1)^{2} + (y + 5)^{2} = 100$

\begin{mfpic}[8]{-12}{10}{-16}{6}

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\axislabels {y}{{$-15$} -15, {$5$} 5}

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\end{mfpic}

\vfill

\columnbreak

\item $(x-4)^2+(y+2)^2 = 9$

\begin{mfpic}[20]{-1}{8}{-6}{2}

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\axislabels {y}{{$-5$} -5, {$-2$} -2, {$1$} 1 }

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(x + 3\right)^{2} + \left(y - \frac{7}{13}\right)^{2} = \frac{1}{4}$

\begin{mfpic}[35]{-4}{1}{-0.75}{2}

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\axislabels {x}{{$-\frac{7}{2} \hspace{6pt}$} -3.5, {$-3 \hspace{6pt}$} -3, {$-\frac{5}{2} \hspace{6pt}$} -2.5}

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\normalsize

\end{mfpic}

\vfill

\columnbreak

\item $(x - 5)^{2} + (y + 9)^{2} = (\ln(8))^{2}$

\begin{mfpic}[10]{-1}{8}{-12}{1}

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\tiny

\axislabels {x}{{$5 - \ln(8)$} 2.92055, {$5$} 5, {$5 + \ln(8)$} 7.07944}

\axislabels {y}{{$-9 - \ln(8)$} -11.07944, {$-9$} -9, {$-9 + \ln(8)$} -6.92055}

\normalsize

\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x + e)^{2} + \left(y - \sqrt{2} \right)^{2} = \pi^{2}$

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\tiny

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\tlabel(0.5,4.55581){$\sqrt{2}+\pi$}

\normalsize

\end{mfpic}

\vfill

\columnbreak

\item $\left(x - \pi \right)^{2} + \left(y - e^{2}\right)^{2} = 91^{\frac{2}{3}}$

\begin{mfpic}[15]{-2}{8.25}{-0.25}{13}

\axes

\circle{(3.14159,7.389),4.4979}

\plotsymbol[3pt]{Cross}{(3.14159,7.389)}

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\ymarks{2.8911, 7.389, 11.88699}

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\tlpointsep{4pt}

\tiny

\axislabels {x}{{$\pi - \sqrt[3]{91}$} -1.3563, {$\pi$} 3.14159, {$\pi + \sqrt[3]{91}$} 7.6395}

\axislabels {y}{{$e^{2} - \sqrt[3]{91}$} 2.8911, {$e^{2}$} 7.389, {$e^{2} + \sqrt[3]{91}$} 11.88699}

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x - 2)^{2} + (y + 5)^{2} = 4$\\

Center $(2, -5)$, radius $r = 2$

\item $(x + 9)^{2} + y^{2} = 25$\\

Center $(-9, 0)$, radius $r = 5$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x+4)^2 + (y-5)^2 = 42$ \\

Center $(-4,5)$, radius $r = \sqrt{42}$

\item $\left(x + \frac{5}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{30}{4}$ \\

Center $\left( -\frac{5}{2}, \frac{1}{2}\right)$, radius $r = \frac{\sqrt{30}}{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^{2} + (y - 3)^{2} = 0$\\

This is not a circle.

\item $\left(x + \frac{1}{2}\right)^{2} + \left(y - \frac{3}{5}\right)^{2} = \frac{161}{100}$\\

Center $\left(-\frac{1}{2}, \frac{3}{5}\right)$, radius $r = \frac{\sqrt{161}}{10}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x - 3)^{2} + (y - 5)^{2} = 65$

\item $(x-3)^2+(y-6)^2 = 20$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x-1)^2 + (y-5)^2 = 5$

\item $(x-1)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{13}{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^{2} + (y - 72)^{2} = 4096$

\end{enumerate}

## 7.3: Parabolas

In Exercises \ref{parabolasketchfirst} - \ref{parabolasketchlast}, sketch the graph of the given parabola. Find the vertex, focus and directrix. Include the endpoints of the latus rectum in your sketch.

\begin{multicols}{2}

\begin{enumerate}

\item $(x - 3)^{2} = -16y$ \label{parabolasketchfirst}

\item $\left(x + \frac{7}{3}\right)^{2} = 2\left(y + \frac{5}{2}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(y - 2)^{2} = -12(x + 3)$

\item $(y + 4)^{2} = 4x$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x-1)^2 = 4(y+3)$

\item $(x+2)^2 = -20(y-5)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(y-4)^2 = 18(x-2)$

\item $\left(y+ \frac{3}{2}\right)^2 = -7 \left(x+ \frac{9}{2}\right)$ \label{parabolasketchlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{stdfrmparabolafirst} - \ref{stdfrmparabolalast}, put the equation into standard form and identify the vertex, focus and directrix.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y^{2} - 10y - 27x + 133 = 0$ \label{stdfrmparabolafirst}

\item $25x^{2} + 20x + 5y - 1 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2 + 2x - 8y + 49 = 0$

\item $2y^2 + 4y +x - 8 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2-10x+12y+1=0$

\item $3y^2-27y+4x+\frac{211}{4} = 0$ \label{stdfrmparabolalast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{buildparafirst} - \ref{buildparalast}, find an equation for the parabola which fits the given criteria.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Vertex $(7, 0)$, focus $(0, 0)$ \label{buildparafirst}

\item Focus $(10, 1)$, directrix $x = 5$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Vertex $(-8, -9)$; $(0, 0)$ and $(-16, 0)$ are points on the curve

\item The endpoints of latus rectum are $(-2, -7)$ and $(4, -7)$ \label{buildparalast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The mirror in Carl's flashlight is a paraboloid of revolution. If the mirror is 5 centimeters in diameter and 2.5 centimeters deep, where should the light bulb be placed so it is at the focus of the mirror?

\item A parabolic Wi-Fi antenna is constructed by taking a flat sheet of metal and bending it into a parabolic shape.\footnote{This shape is called a parabolic cylinder.'} If the cross section of the antenna is a parabola which is 45 centimeters wide and 25 centimeters deep, where should the receiver be placed to maximize reception?

\item \label{parabolaarch} A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch.

\item A popular novelty item is the mirage bowl.' Follow this \href{http://spie.org/etop/2007/etop07meth...derline{link}} to see another startling application of the reflective property of the parabola.

\item With the help of your classmates, research spinning liquid mirrors. To get you started, check out this \href{http://www.astro.ubc.ca/LMT/lzt/}{\u...line{website}}.

\end{enumerate}

\newpage

\begin{enumerate}

\item \begin{multicols}{2}

{\small $(x - 3)^{2} = -16y$}\\

{\small Vertex $(3, 0)$}\\

{\small Focus $(3, -4)$}\\

{\small Directrix $y = 4$}\\

{\small Endpoints of latus rectum $(-5, -4)$, $(11, -4)$}\\

\vfill

\columnbreak

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\end{multicols}

\smallskip

\item \begin{multicols}{2}

{\small $\left(x + \frac{7}{3}\right)^{2} = 2\left(y + \frac{5}{2}\right)$}\\

{\small Vertex $\left(-\frac{7}{3}, -\frac{5}{2} \right)$}\\

{\small Focus $\left(-\frac{7}{3}, -2 \right)$}\\

{\small Directrix $y = -3$}\\

{\small Endpoints of latus rectum $\left(-\frac{10}{3}, -2 \right)$, $\left(-\frac{4}{3}, -2 \right)$}\\

\vfill

\columnbreak

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\smallskip

\item \begin{multicols}{2}

{\small $(y - 2)^{2} = -12(x + 3)$} \\

{\small Vertex $(-3, 2)$} \\

{\small Focus $(-6, 2)$} \\

{\small Directrix $x = 0$}\\

{\small Endpoints of latus rectum $(-6, 8)$, $(-6, -4)$}\\

\vfill

\columnbreak

\begin{mfpic}[10]{-8}{1}{-5}{9}

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\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2}

{\small $(y + 4)^{2} = 4x$}\\

{\small Vertex $(0,-4)$} \\

{\small Focus $(1,-4)$} \\

{\small Directrix $x = -1$}\\

{\small Endpoints of latus rectum $(1, -2)$, $(1, -6)$}\\

\vfill

\columnbreak

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\end{mfpic}

\end{multicols}

\smallskip

\item \begin{multicols}{2}

{\small $(x-1)^2 = 4(y+3)$}\\

{\small Vertex $\left(1, -3\right)$}\\

{\small Focus $\left(1, -2 \right)$}\\

{\small Directrix $y = -4$}\\

{\small Endpoints of latus rectum $\left(3, -2 \right)$, $\left(-1, -2 \right)$}\\

\vfill

\columnbreak

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\smallskip

\item \begin{multicols}{2}

{\small $(x+2)^2 = -20(y-5)$}\\

{\small Vertex $\left(-2, 5\right)$}\\

{\small Focus $\left(-2, 0 \right)$}\\

{\small Directrix $y = 10$}\\

{\small Endpoints of latus rectum $\left(-12, 0 \right)$, $\left(8, 0 \right)$}\\

\vfill

\columnbreak

\begin{mfpic}[7.5][10]{-13}{9}{-1}{11}

\axes

\xmarks{-12 step 1 until 8}

\ymarks{1 step 1 until 10}

\arrow \reverse \arrow \function{-13,9,0.1}{((x +2)**2)/(0-20) + 5}

\arrow \reverse \arrow \polyline{(-13,10),(9,10)}

\plotsymbol[3pt]{Asterisk}{(-2,0)}

\tlabel(9,-0.5){\scriptsize $x$}

\tlabel(0.5,11){\scriptsize $y$}

\point[3pt]{(-12,0),(-2,5),(8,0)}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-12 \hspace{7pt}$} -12, {$-10 \hspace{7pt}$} -10, {$-8 \hspace{7pt}$} -8, {$-6 \hspace{7pt}$} -6, {$-4 \hspace{7pt}$} -4, {$-2 \hspace{7pt}$} -2, {$2$} 2, {$4$} 4, {$6$} 6, {$8$} 8}

\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9, {$10$} 10}

\normalsize

\end{mfpic}

\end{multicols}

\smallskip

\item \begin{multicols}{2}

{\small $(y-4)^2 = 18(x-2)$}\\

{\small Vertex $\left(2, 4\right)$}\\

{\small Focus $\left( \frac{13}{2}, 4 \right)$}\\

{\small Directrix $x = -\frac{5}{2}$}\\

{\small Endpoints of latus rectum $\left(\frac{13}{2}, -5 \right)$, $\left(\frac{13}{2}, 13 \right)$}\\

\vfill

\columnbreak

\begin{mfpic}[15][7.5]{-3}{8}{-6}{14}

\axes

\xmarks{-2 step 1 until 7}

\ymarks{-5 step 1 until 13}

\arrow \function{2,8,0.1}{4+(sqrt(18*(x-2)))}

\arrow \function{2,8,0.1}{4-(sqrt(18*(x-2)))}

\arrow \reverse \arrow \polyline{(-2.5,-6),(-2.5,14)}

\plotsymbol[3pt]{Asterisk}{(6.5,4)}

\tlabel(8,-0.5){\scriptsize $x$}

\tlabel(0.5,14){\scriptsize $y$}

\point[3pt]{(6.5,-5),(2,4),(6.5,13)}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7}

\axislabels {y}{{$-5$} -5, {$-3$} -3, {$-1$} -1, {$1$} 1, {$3$} 3, {$5$} 5, {$7$} 7, {$9$} 9, {$11$} 11, {$13$} 13}

\normalsize

\end{mfpic}

\end{multicols}

\smallskip

\item \begin{multicols}{2}

{\small $\left(y+ \frac{3}{2}\right)^2 = -7 \left(x+ \frac{9}{2}\right)$}\\

{\small Vertex $\left(-\frac{9}{2}, -\frac{3}{2}\right)$}\\

{\small Focus $\left( -\frac{25}{4}, -\frac{3}{2} \right)$}\\

{\small Directrix $x = -\frac{11}{4}$}\\

{\small Endpoints of latus rectum $\left(-\frac{25}{4}, 2 \right)$, $\left(-\frac{25}{4}, -5 \right)$}\\

\vfill

\columnbreak

\begin{mfpic}[15]{-7}{1}{-6}{3}

\axes

\xmarks{-6 step 1 until -1}

\ymarks{-5 step 1 until 2}

\arrow \function{-4.5,-7,0.1}{0-1.5+(sqrt((0-7)*(x+4.5)))}

\arrow \function{-4.5,-7,0.1}{0-1.5-(sqrt((0-7)*(x+4.5)))}

\arrow \reverse \arrow \polyline{(-2.75,-6),(-2.75,3)}

\plotsymbol[3pt]{Asterisk}{(-6.25,-1.5)}

\tlabel(1,-0.5){\scriptsize $x$}

\tlabel(0.5,3){\scriptsize $y$}

\point[3pt]{(-6.25,-5),(-4.5,-1.5),(-6.25,2)}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-5 \hspace{7pt}$} -5,{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2,{$-1 \hspace{7pt}$} -1}

\axislabels {y}{{$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2}

\normalsize

\end{mfpic}

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(y - 5)^{2} = 27(x - 4)$\\

Vertex $(4, 5)$\\

Focus $\left( \frac{43}{4}, 5 \right)$\\

Directrix $x = -\frac{11}{4}$

\item $\left(x + \frac{2}{5} \right)^{2} = -\frac{1}{5}(y - 1)$\\

Vertex $\left( -\frac{2}{5}, 1 \right)$\\

Focus $\left( -\frac{2}{5}, \frac{19}{20} \right)$\\

Directrix $y = \frac{21}{20}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x+1)^2=8(y-6)$ \\

Vertex $(-1,6)$\\

Focus $(-1,8)$ \\

Directrix $y=4$

\item $(y+1)^2=-\frac{1}{2}(x-10)$\\

Vertex $(10,-1)$\\

Focus $\left(\frac{79}{8}, -1 \right)$\\

Directrix $x = \frac{81}{8}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x-5)^2 = -12(y-2)$\\

Vertex $(5,2)$\\

Focus $(5,-1)$ \\

Directrix $y=5$

\item $\left(y-\frac{9}{2}\right)^2 = -\frac{4}{3} (x-2)$\\

Vertex $\left(2, \frac{9}{2}\right)$\\

Focus $\left(\frac{5}{3}, \frac{9}{2}\right)$\\

Directrix $x = \frac{7}{3}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y^{2} = -28(x - 7)$

\item $(y - 1)^{2} = 10\left(x - \frac{15}{2} \right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x + 8)^{2} = \frac{64}{9}(y + 9)$

\item $(x - 1)^{2} = 6\left(y + \frac{17}{2}\right)$ or\\

$(x - 1)^{2} = -6\left(y + \frac{11}{2}\right)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The bulb should be placed $0.625$ centimeters above the vertex of the mirror. (As verified by Carl himself!)

\item The receiver should be placed $5.0625$ centimeters from the vertex of the cross section of the antenna.

\item The arch can be modeled by $x^2=-(y-9)$ or $y=9-x^2$. One foot in from the base of the arch corresponds to either $x = \pm 2$, so the height is $y=9-(\pm 2)^2=5$ feet.

\end{enumerate}

\closegraphsfile

## 7.4: Ellipses

\subsection{Exercises}

In Exercises \ref{graphellipseexfirst} - \ref{graphellipseexlast}, graph the ellipse. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, the foci and the eccentricity.

\begin{multicols}{2}

\begin{enumerate}

\item $\dfrac{x^{2}}{169} + \dfrac{y^{2}}{25} = 1$ \label{graphellipseexfirst}

\item $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 2)^{2}}{4} + \dfrac{(y + 3)^{2}}{9} = 1$

\item $\dfrac{(x + 5)^{2}}{16} + \dfrac{(y - 4)^{2}}{1} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 1)^{2}}{10} + \dfrac{(y - 3)^{2}}{11} = 1$

\item $\dfrac{(x-1)^2}{9}+\dfrac{(y+3)^2}{4} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x+2)^2}{16}+\dfrac{(y-5)^2}{20} = 1$

\item $\dfrac{(x-4)^2}{8} + \dfrac{(y-2)^2}{18} = 1$ \label{graphellipseexlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{stdformellipseexfirst} - \ref{stdformellipseexlast}, put the equation in standard form. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, the foci and the eccentricity.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $9x^2+25y^2-54x-50y-119=0$ \label{stdformellipseexfirst}

\item $12x^{2} + 3y^{2} - 30y + 39 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $5x^{2} + 18y^{2} - 30x + 72y + 27 = 0$

\item $x^2 - 2x + 2y^2 - 12y + 3 = 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $9x^2 + 4y^2 - 4y - 8 = 0$

\item $6x^2+5y^2-24x+20y+14=0$ \label{stdformellipseexlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{buildellipsefirst} - \ref{buildellipselast}, find the standard form of the equation of the ellipse which has the given properties.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Center $(3, 7)$, Vertex $(3, 2)$, Focus $(3, 3)$ \label{buildellipsefirst}

\item Foci $(0, \pm 5)$, Vertices $(0, \pm 8)$.

\item Foci $(\pm 3, 0)$, length of the Minor Axis $10$

\item Vertices $(3,2)$, $(13,2)$; Endpoints of the Minor Axis $(8,4)$, $(8,0)$

\item Center $(5,2)$, Vertex $(0,2)$, eccentricity $\frac{1}{2}$

\item All points on the ellipse are in Quadrant IV except $(0, -9)$ and $(8, 0)$. (One might also say that the ellipse is tangent to the axes'' at those two points.) \label{buildellipselast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Repeat Example \ref{whisgalleryex} for a whispering gallery 200 feet wide and 75 feet tall.

\item \label{ellipsearchex} An elliptical arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch. Compare your result with your answer to Exercise \ref{parabolaarch} in Section \ref{Parabolas}.

\item The Earth's orbit around the sun is an ellipse with the sun at one focus and eccentricity $e \approx 0.0167$. The length of the semimajor axis (that is, half of the major axis) is defined to be $1$ astronomical unit (AU). The vertices of the elliptical orbit are given special names: aphelion' is the vertex farthest from the sun, and perihelion' is the vertex closest to the sun. Find the distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion.

\item The graph of an ellipse clearly fails the Vertical Line Test, Theorem \ref{VLT}, so the equation of an ellipse does not define $y$ as a function of $x$. However, much like with circles and horizontal parabolas, we can split an ellipse into a top half and a bottom half, each of which would indeed represent $y$ as a function of $x$. With the help of your classmates, use your calculator to graph the ellipses given in Exercises \ref{graphellipseexfirst} - \ref{graphellipseexlast} above. What difficulties arise when you plot them on the calculator?

\item Some famous examples of whispering galleries include \href{http://www.stpauls.co.uk/}{\underline{St. Paul's Cathedral}} in London, England, \href{http://www.aoc.gov/cc/capitol/nat_st...rline{National Statuary Hall}} in Washington, D.C., and \href{http://www.cincymuseum.org/}{\underline{The Cincinnati Museum Center}}. With the help of your classmates, research these whispering galleries. How does the whispering effect compare and contrast with the scenario in Example \ref{whisgalleryex}?

\item With the help of your classmates, research extracorporeal shock-wave lithotripsy''. It uses the reflective property of the ellipsoid to dissolve kidney stones.

\end{enumerate}

\newpage

\begin{enumerate}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{x^{2}}{169} + \dfrac{y^{2}}{25} = 1$

Center $(0, 0)$\\

Major axis along $y = 0$\\

Minor axis along $x = 0$\\

Vertices $(13, 0), \, (-13, 0)$\\

Endpoints of Minor Axis $(0,-5)$, $(0,5)$ \\

Foci $(12, 0), \, (-12, 0)$\\

$e = \frac{12}{13}$\\

\begin{mfpic}[7][10]{-14}{14}{-6}{6}

\axes

\tlabel(14,-0.5){\scriptsize $x$}

\tlabel(0.5,6){\scriptsize $y$}

\xmarks{-13 step 1 until 13}

\ymarks{-5 step 1 until 5}

\ellipse{(0,0),13,5}

\plotsymbol[3pt]{Asterisk}{(-12,0), (12,0)}

\plotsymbol[3pt]{Cross}{(0,0)}

\point[3pt]{(-13,0), (13,0), (0,5), (0,-5)}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-13 \hspace{6pt}$} -13, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$13$} 13}

\axislabels {y}{{$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{x^{2}}{9} + \dfrac{y^{2}}{25} = 1$

Center $(0, 0)$\\

Major axis along $x = 0$\\

Minor axis along $y = 0$\\

Vertices $(0,5), \, (0,-5)$\\

Endpoints of Minor Axis $(-3,0)$, $(3,0)$ \\

Foci $(0,-4), \, (0,4)$\\

$e = \frac{4}{5}$\\

\begin{mfpic}[10]{-4}{4}{-6}{6}

\axes

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,6){\scriptsize $y$}

\xmarks{-3 step 1 until 3}

\ymarks{-5 step 1 until 5}

\ellipse{(0,0),3,5}

\plotsymbol[3pt]{Asterisk}{(0,-4), (0,4)}

\plotsymbol[3pt]{Cross}{(0,0)}

\point[3pt]{(-3,0), (3,0), (0,5), (0,-5)}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1,{$1$} 1, {$2$} 2, {$3$} 3}

\axislabels {y}{{$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x - 2)^{2}}{4} + \dfrac{(y + 3)^{2}}{9} = 1$

Center $(2, -3)$\\

Major axis along $x = 2$\\

Minor axis along $y = -3$\\

Vertices $(2, 0), \, (2, -6)$\\

Endpoints of Minor Axis $(0,-3)$, $(4,-3)$\\

Foci $(2, -3 + \sqrt{5}), \, (2, -3 - \sqrt{5})$\\

$e = \frac{\sqrt{5}}{3}$\\

\begin{mfpic}[15]{-1}{5}{-7}{1}

\axes

\tlabel(5,-0.5){\scriptsize $x$}

\tlabel(0.5,1){\scriptsize $y$}

\xmarks{1 step 1 until 4}

\ymarks{-6 step 1 until 0}

\ellipse{(2,-3),2,3}

\plotsymbol[3pt]{Asterisk}{(2, -0.7639), (2,-5.23606)}

\plotsymbol[3pt]{Cross}{(2,-3)}

\point[3pt]{(2,0), (2,-6), (0,-3), (4,-3)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x + 5)^{2}}{16} + \dfrac{(y - 4)^{2}}{1} = 1$

Center $(-5, 4)$\\

Major axis along $y = 4$\\

Minor axis along $x = -5$\\

Vertices $(-9, 4), \, (-1, 4)$\\

Endpoints of Minor Axis $(-5,3)$, $(-5,5)$\\

Foci $(-5 + \sqrt{15}, 4), \, (-5 - \sqrt{15}, 4)$\\

$e = \frac{\sqrt{15}}{4}$\\

\begin{mfpic}[16]{-10}{1}{-.5}{6}

\axes

\tlabel(1,-0.5){\scriptsize $x$}

\tlabel(0.5,6){\scriptsize $y$}

\xmarks{-9 step 1 until -1}

\ymarks{1 step 1 until 5}

\ellipse{(-5,4),4,1}

\plotsymbol[3pt]{Asterisk}{(-8.873,4), (-1.127,4)}

\plotsymbol[3pt]{Cross}{(-5,4)}

\point[3pt]{(-9,4), (-1,4), (-5,5), (-5,3)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-9 \hspace{7pt}$} -9, {$-8 \hspace{7pt}$} -8, {$-7 \hspace{7pt}$} -7, {$-6 \hspace{7pt}$} -6, {$-5 \hspace{7pt}$} -5, {$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1}

\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x - 1)^{2}}{10} + \dfrac{(y - 3)^{2}}{11} = 1$

Center $(1, 3)$\\

Major axis along $x = 1$\\

Minor axis along $y = 3$\\

Vertices $(1, 3 + \sqrt{11}), \, (1, 3 - \sqrt{11})$\\

Endpoints of the Minor Axis \\ $(1-\sqrt{10}, 3), \, (1+\sqrt{10}, 3)$\\

Foci $(1, 2), \, (1, 4)$\\

$e = \frac{\sqrt{11}}{11}$\\

\begin{mfpic}[18]{-3}{5}{-1}{7}

\axes

\tlabel(5,-0.25){\scriptsize $x$}

\tlabel(0.25,7){\scriptsize $y$}

\xmarks{-2 step 1 until 4}

\ymarks{1 step 1 until 6}

\ellipse{(1,3),3.1623,3.3166}

\plotsymbol[3pt]{Asterisk}{(1,2), (1,4)}

\plotsymbol[3pt]{Cross}{(1,3)}

\point[3pt]{(1,6.3166), (1,-0.3166), (-2.1623, 3), (4.1623,3)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x-1)^2}{9}+\dfrac{(y+3)^2}{4} = 1$

Center $(1, -3)$\\

Major axis along $y = -3$\\

Minor axis along $x = 1$\\

Vertices $(4, -3), \, (-2, -3)$\\

Endpoints of the Minor Axis $(1,-1)$, $(1,-5)$\\

Foci $(1+\sqrt{5}, -3), \, (1-\sqrt{5}, -3)$\\

$e = \frac{\sqrt{5}}{3}$\\

\begin{mfpic}[18]{-3}{5}{-6}{1}

\axes

\tlabel(5,-0.25){\scriptsize $x$}

\tlabel(0.25,1){\scriptsize $y$}

\xmarks{-2 step 1 until 4}

\ymarks{-5 step 1 until -1}

\ellipse{(1,-3),3,2}

\plotsymbol[3pt]{Asterisk}{(3.2361,-3), (-1.2361,-3)}

\plotsymbol[3pt]{Cross}{(1,-3)}

\point[3pt]{(4,-3), (-2,-3), (1,-1), (1,-5)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\axislabels {y}{{$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x+2)^2}{16}+\dfrac{(y-5)^2}{20} = 1$

Center $(-2, 5)$\\

Major axis along $x = -2$\\

Minor axis along $y = 5$\\

Vertices $(-2, 5 + 2\sqrt{5}), \, (-2, 5 - 2\sqrt{5})$\\

Endpoints of the Minor Axis $(-6,5)$, $(2,5)$

Foci $(-2, 7), \, (-2, 3)$\\

$e = \frac{\sqrt{5}}{5}$\\

\begin{mfpic}[18]{-7}{3}{-1}{11}

\axes

\tlabel(3,-0.25){\scriptsize $x$}

\tlabel(0.25,11){\scriptsize $y$}

\xmarks{-6 step 1 until 2}

\ymarks{1 step 1 until 11}

\ellipse{(-2,5),4,4.4721}

\plotsymbol[3pt]{Asterisk}{(-2,7), (-2,3)}

\plotsymbol[3pt]{Cross}{(-2,5)}

\point[3pt]{(-2,9.4721), (-2,0.5279), (-6,5), (2,5)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$-6 \hspace{7pt}$} -6,{$-5 \hspace{7pt}$} -5,{$-4 \hspace{7pt}$} -4,{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}

\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9, {$10$} 10 }

\normalsize

\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x-4)^2}{8}+\dfrac{(y-2)^2}{18} = 1$

Center $(4, 2)$\\

Major axis along $x = 4$\\

Minor axis along $y = 2$\\

Vertices $(4, 2 + 3\sqrt{2}), \, (4, 2 - 3\sqrt{2})$\\

Endpoints of the Minor Axis \\

$(4-2\sqrt{2},2)$, $(4+2\sqrt{2},2)$\\

Foci $(4, 2+\sqrt{10}), \, (4, 2-\sqrt{10})$\\

$e = \frac{\sqrt{5}}{3}$\\

\begin{mfpic}[18]{-1}{8}{-4}{8}

\axes

\tlabel(8,-0.25){\scriptsize $x$}

\tlabel(0.25,8){\scriptsize $y$}

\xmarks{1 step 1 until 7}

\ymarks{-3 step 1 until 7}

\ellipse{(4,2),2.8284,4.2426}

\plotsymbol[3pt]{Asterisk}{(4,5.1623), (4,-1.1623)}

\plotsymbol[3pt]{Cross}{(4,2)}

\point[3pt]{(4,6.2426), (4,-2.2426), (1.1716, 2), (6.828,2)}

\tlpointsep{4pt}

\scriptsize

\axislabels {x}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7}

\axislabels {y}{{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7}

\normalsize

\end{mfpic}

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x-3)^2}{25} + \dfrac{\left(y-1\right)^2}{9} = 1$\\

Center $\left(3, 1 \right)$\\

Major Axis along $y=1$\\

Minor Axis along $x=3$\\

Vertices $\left( 8, 1 \right)$, $(-2, 1)$\\

Endpoints of Minor Axis $\left(3,4\right)$, $\left(3,-2\right)$\\

Foci $\left(7,1 \right)$, $\left(-1, 1\right)$\\

$e = \frac{4}{5}$

\vfill

\columnbreak

\item $\dfrac{x^{2}}{3} + \dfrac{(y - 5)^{2}}{12} = 1$\\

Center $(0, 5)$\\

Major axis along $x = 0$\\

Minor axis along $y = 5$\\

Vertices $(0, 5 - 2\sqrt{3}), (0, 5 + 2\sqrt{3})$\\

Endpoints of Minor Axis $(-\sqrt{3},5)$, $(\sqrt{3},5)$\\

Foci $(0, 2), (0, 8)$\\

$e = \frac{\sqrt{3}}{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 3)^{2}}{18} + \dfrac{(y + 2)^{2}}{5} = 1$\\

Center $(3, -2)$\\

Major axis along $y = -2$\\

Minor axis along $x = 3$\\

Vertices $(3 - 3\sqrt{2}, -2), (3 + 3\sqrt{2}, -2)$\\

Endpoints of Minor Axis $(3,-2+\sqrt{5})$, $(3,-2-\sqrt{5})$\\

Foci $(3 - \sqrt{13}, -2), (3 + \sqrt{13}, -2)$\\

$e = \frac{\sqrt{26}}{6}$

\vfill

\columnbreak

\item $\dfrac{(x - 1)^{2}}{16} + \dfrac{(y - 3)^{2}}{8} = 1$\\

Center $(1,3)$ \\

Major Axis along $y=3$\\

Minor Axis along $x=1$\\

Vertices $(5, 3)$, $(-3,3)$\\

Endpoints of Minor Axis $(1,3+2\sqrt{2})$, $(1,3-2\sqrt{2})$\\

Foci $(1 + 2 \sqrt{2}, 3)$, $(1-2 \sqrt{2},3)$\\

$e = \frac{\sqrt{2}}{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{x^2}{1} + \dfrac{4\left(y-\frac{1}{2}\right)^2}{9} = 1$\\

Center $\left(0, \frac{1}{2} \right)$\\

Major Axis along $x=0$ (the $y$-axis)\\

Minor Axis along $y=\frac{1}{2}$\\

Vertices $\left( 0, 2 \right)$, $(0, -1)$\\

Endpoints of Minor Axis $\left(-1, \frac{1}{2} \right)$, $\left(1, \frac{1}{2} \right)$\\

Foci $\left(0, \frac{1+\sqrt{5}}{2}\right)$, $\left(0, \frac{1-\sqrt{5}}{2}\right)$\\

$e = \frac{\sqrt{5}}{3}$

\vfill

\columnbreak

\item $\dfrac{(x-2)^2}{5} + \dfrac{\left(y+2\right)^2}{6} = 1$\\

Center $\left(2, -2 \right)$\\

Major Axis along $x=2$\\

Minor Axis along $y=-2$\\

Vertices $\left( 2, -2+\sqrt{6} \right)$, $(2, -2-\sqrt{6})$\\

Endpoints of Minor Axis $\left(2-\sqrt{5},-2 \right)$, $\left(2+\sqrt{5},-2\right)$\\

Foci $\left(2,-1 \right)$, $\left(2, -3\right)$\\

$e = \frac{\sqrt{6}}{6}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 3)^{2}}{9} + \dfrac{(y - 7)^{2}}{25} = 1$

\item $\dfrac{x^{2}}{39} + \dfrac{y^{2}}{64} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{x^{2}}{34} + \dfrac{y^{2}}{25} = 1$

\item $\dfrac{(x - 8)^{2}}{25} + \dfrac{(y - 2)^{2}}{4} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x-5)^{2}}{25} + \dfrac{4(y-2)^{2}}{75} = 1$

\item $\dfrac{(x - 8)^{2}}{64} + \dfrac{(y + 9)^{2}}{81} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Jamie and Jason should stand $100-25\sqrt{7} \approx 33.86$ feet from opposite ends of the gallery.

\item The arch can be modeled by the top half of $\frac{x^2}{9} + \frac{y^2}{81} = 1$. One foot in from the base of the arch corresponds to either $x = \pm 2$. Plugging in $x = \pm 2$ gives $y = \pm 3\sqrt{5}$ and since $y$ represents a height, we choose $y=3\sqrt{5} \approx 6.71$ feet.

\item Distance from the sun to aphelion $\approx 1.0167$ AU.\\

Distance from the sun to perihelion $\approx 0.9833$ AU.

\end{enumerate}

\closegraphsfile

## 7.5: Hyperbolas

\subsection{Exercises}

In Exercises \ref{graphhyperbolafirst} - \ref{graphhyperbolalast}, graph the hyperbola. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes.

\begin{multicols}{2}

\begin{enumerate}

\item $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$ \label{graphhyperbolafirst}

\item $\dfrac{y^{2}}{9} - \dfrac{x^{2}}{16} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 2)^{2}}{4} - \dfrac{(y + 3)^{2}}{9} = 1$

\item $\dfrac{(y - 3)^{2}}{11} - \dfrac{(x - 1)^{2}}{10} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x + 4)^{2}}{16} - \dfrac{(y - 4)^{2}}{1} = 1$

\item $\dfrac{(x+1)^2}{9} - \dfrac{(y-3)^2}{4} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(y+2)^2}{16} - \dfrac{(x-5)^2}{20} = 1$

\item $\dfrac{(x-4)^2}{8} - \dfrac{(y-2)^2}{18} = 1$ \label{graphhyperbolalast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{stdfrmhypfirst} - \ref{stdfrmhyplast}, put the equation in standard form. Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $12x^{2} - 3y^{2} + 30y - 111 = 0$ \label{stdfrmhypfirst}

\item $18y^{2} - 5x^{2} + 72y + 30x - 63= 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $9x^2-25y^2-54x-50y-169 = 0$

\item $-6x^2+5y^2-24x+40y+26=0$ \label{stdfrmhyplast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{buildhypfirst} - \ref{buildhyplast}, find the standard form of the equation of the hyperbola which has the given properties.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Center $(3, 7)$, Vertex $(3, 3)$, Focus $(3, 2)$ \label{buildhypfirst}

\item Vertex $(0, 1)$, Vertex $(8, 1)$, Focus $(-3, 1)$

\item Foci $(0, \pm 8)$, Vertices $(0, \pm 5)$.

\item Foci $(\pm 5, 0)$, length of the Conjugate Axis $6$

\item Vertices $(3,2)$, $(13,2)$; Endpoints of the Conjugate Axis $(8,4)$, $(8,0)$

\item Vertex $(-10, 5)$, Asymptotes $y = \pm \frac{1}{2}(x - 6) + 5$ \label{buildhyplast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{generalconicfirst} - \ref{generalconiclast}, find the standard form of the equation using the guidelines on page \pageref{idconocsrulesofthumb} and then graph the conic section.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x^2-2x-4y-11=0$ \label{generalconicfirst}

\item $x^2 + y^2-8x+4y+11=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $9x^2 + 4y^2-36x+24y + 36=0$

\item $9x^2-4y^2-36x-24y-36=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $y^2+8y-4x+16=0$

\item $4x^2+y^2-8x+4=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $4x^2+9y^2-8x+54y+49=0$

\item $x^2 + y^2-6x+4y+14=0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $2x^2+ 4y^2+12x-8y+25=0$

\item $4x^2-5y^2-40x-20y+160=0$ \label{generalconiclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The graph of a vertical or horizontal hyperbola clearly fails the Vertical Line Test, Theorem \ref{VLT}, so the equation of a vertical of horizontal hyperbola does not define $y$ as a function of $x$.\footnote{We will see later in the text that the graphs of certain rotated hyperbolas pass the Vertical Line Test.} However, much like with circles, horizontal parabolas and ellipses, we can split a hyperbola into pieces, each of which would indeed represent $y$ as a function of $x$. With the help of your classmates, use your calculator to graph the hyperbolas given in Exercises \ref{graphhyperbolafirst} - \ref{graphhyperbolalast} above. How many pieces do you need for a vertical hyperbola? How many for a horizontal hyperbola?

\item The location of an earthquake's epicenter $-$ the point on the surface of the Earth directly above where the earthquake actually occurred $-$ can be determined by a process similar to how we located Sasquatch in Example \ref{FindtheSasquatch}. (As we said back in Exercise \ref{Richterexercise} in Section \ref{IntroExpLogs}, earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a course in Geology or the U.S. Geological Survey's Earthquake Hazards Program found \href{http://earthquake.usgs.gov/}{\underline{here}}.) Our technique works only for relatively small distances because we need to assume that the Earth is flat in order to use hyperbolas in the plane.\footnote{Back in the Exercises in Section \ref{CartesianPlane} you were asked to research people who believe the world is flat. What did you discover?} The P-waves (P'' stands for Primary) of an earthquake in Sasquatchia travel at 6 kilometers per second.\footnote{Depending on the composition of the crust at a specific location, P-waves can travel between 5 kps and 8 kps.} Station A records the waves first. Then Station B, which is 100 kilometers due north of Station A, records the waves 2 seconds later. Station C, which is 150 kilometers due west of Station A records the waves 3 seconds after that (a total of 5 seconds after Station A). Where is the epicenter?

\item \label{hyperbolaeccentricity} The notion of eccentricity introduced for ellipses in Definition \ref{ellipseeccentricity} in Section \ref{Ellipses} is the same for hyperbolas in that we can define the eccentricity $e$ of a hyperbola as

$e = \dfrac{\mbox{distance from the center to a focus}}{\mbox{distance from the center to a vertex}}$

\begin{enumerate}

\item With the help of your classmates, explain why $e > 1$ for any hyperbola.

\item Find the equation of the hyperbola with vertices $(\pm 3,0)$ and eccentricity $e = 2$.

\item With the help of your classmates, find the eccentricity of each of the hyperbolas in Exercises \ref{graphhyperbolafirst} - \ref{graphhyperbolalast}. What role does eccentricity play in the shape of the graphs?

\end{enumerate}

\item On page \pageref{paraboloid} in Section \ref{Parabolas}, we discussed paraboloids of revolution when studying the design of satellite dishes and parabolic mirrors. In much the same way, `natural draft' cooling towers are often shaped as \index{hyperboloid} \textbf{hyperboloids of revolution}. Each vertical cross section of these towers is a hyperbola. Suppose the a natural draft cooling tower has the cross section below. Suppose the tower is 450 feet wide at the base, 275 feet wide at the top, and 220 feet at its narrowest point (which occurs 330 feet above the ground.) Determine the height of the tower to the nearest foot.

\begin{center}

\begin{mfpic}[20]{-3}{3}{0}{5}

\curve{(3,0), (1.5,3), (2,5)}

\curve{(-3,0), (-1.5,3), (-2,5)}

\point[3pt]{(3,0), (1.5,3), (2,5), (-3,0), (-1.5,3), (-2,5)}

\arrow \reverse \arrow \polyline{(-2.75,0), (2.75,0)}

\tlabel[cc](0,-0.5){\scriptsize $450$ ft}

\arrow \reverse \arrow \polyline{(-1.25,3), (1.25,3)}

\tlabel[cc](0,2.5){\scriptsize $220$ ft}

\arrow \reverse \arrow \polyline{(-1.75,5), (1.75,5)}

\tlabel[cc](0,5.5){\scriptsize $275$ ft}

\arrow \reverse \arrow \polyline{(5,0.25), (5,2.75)}

\dotted \polyline{(1.5,3), (5,3)}

\gclear \tlabelrect[cc]{(5,1.5)}{\scriptsize $330$ ft}

\end{mfpic}

\end{center}

\item With the help of your classmates, research the Cassegrain Telescope. It uses the reflective property of the hyperbola as well as that of the parabola to make an ingenious telescope.

\item \label{conicsclassificationnoxytermex} With the help of your classmates show that if $Ax^2 + Cy^2 + Dx + Ey + F = 0$ determines a non-degenerate conic\footnote{Recall that this means its graph is either a circle, parabola, ellipse or hyperbola.} then

\begin{itemize}

\item $AC < 0$ means that the graph is a hyperbola

\item $AC = 0$ means that the graph is a parabola

\item $AC > 0$ means that the graph is an ellipse or circle

\end{itemize}

\textbf{NOTE:} This result will be generalized in Theorem \ref{conicclassification} in Section \ref{rotationaxes}.

\end{enumerate}

\newpage

\begin{enumerate}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$

Center $(0, 0)$\\

Transverse axis on $y = 0$\\

Conjugate axis on $x = 0$\\

Vertices $(4, 0), (-4, 0)$\\

Foci $(5, 0), (-5, 0)$\\

Asymptotes $y = \pm \frac{3}{4} x$\\

\begin{mfpic}[12][9]{-7}{7}{-7}{7}

\axes

\tlabel(7,-0.5){\scriptsize $x$}

\tlabel(0.5,7){\scriptsize $y$}

\xmarks{-6 step 1 until 6}

\ymarks{-6 step 1 until 6}

\point[3pt]{(4,0),(-4,0)}

\dotted[1pt, 3pt] \polyline{(-4,3), (4,3), (4, -3), (-4,-3), (-4,3)}

\arrow \reverse \arrow \parafcn{-5,5,0.1}{(sqrt(16 + (1.778*(t**2))),t)}

\arrow \reverse \arrow \parafcn{-5,5,0.1}{(-sqrt(16 + (1.778*(t**2))),t)}

\arrow \reverse \arrow \dashed \function{-7,7,0.1}{0.75*x}

\arrow \reverse \arrow \dashed \function{-7,7,0.1}{-0.75*x}

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\plotsymbol[3pt]{Asterisk}{(5,0), (-5,0)}

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\axislabels {y}{{$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{y^{2}}{9} - \dfrac{x^{2}}{16} = 1$

Center $(0, 0)$\\

Transverse axis on $x = 0$\\

Conjugate axis on $y = 0$\\

Vertices $(0, 3), (0, -3)$\\

Foci $(0, 5), (0, -5)$\\

Asymptotes $y = \pm \frac{3}{4} x$\\

\begin{mfpic}[12][9]{-7}{7}{-7}{7}

\axes

\tlabel(7,-0.5){\scriptsize $x$}

\tlabel(0.5,7){\scriptsize $y$}

\xmarks{-6 step 1 until 6}

\ymarks{-6 step 1 until 6}

\point[3pt]{(0,3),(0,-3)}

\dotted[1pt, 3pt] \polyline{(-4,3), (4,3), (4, -3), (-4,-3), (-4,3)}

\arrow \reverse \arrow \function{-7,7,0.1}{sqrt(9 + (0.5625*(x**2)))}

\arrow \reverse \arrow \function{-7,7,0.1}{-sqrt(9 + (0.5625*(x**2)))}

\arrow \reverse \arrow \dashed \function{-7,7,0.1}{0.75*x}

\arrow \reverse \arrow \dashed \function{-7,7,0.1}{-0.75*x}

\plotsymbol[3pt]{Cross}{(0,0)}

\plotsymbol[3pt]{Asterisk}{(0,5), (0,-5)}

\tlpointsep{4pt}

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\axislabels {x}{{$-6 \hspace{6pt}$} -6, {$-5 \hspace{6pt}$} -5, {$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6}

\axislabels {y}{{$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x - 2)^{2}}{4} - \dfrac{(y + 3)^{2}}{9} = 1$

Center $(2, -3)$\\

Transverse axis on $y = -3$\\

Conjugate axis on $x = 2$\\

Vertices $(0, -3), (4, -3)$\\

Foci $(2 + \sqrt{13}, -3), (2 - \sqrt{13}, -3)$\\

Asymptotes $y = \pm \frac{3}{2}(x - 2) - 3$\\

\begin{mfpic}[12][9]{-4}{8}{-11}{5}

\axes

\tlabel(8,-0.5){\scriptsize $x$}

\tlabel(0.5,5){\scriptsize $y$}

\xmarks{-3 step 1 until 7}

\ymarks{-10 step 1 until 4}

\point[3pt]{(0,-3),(4,-3)}

\dotted[1pt, 3pt] \polyline{(0,0), (4,0), (4, -6), (0,-6), (0,0)}

\arrow \function{4,7,0.1}{-3 + sqrt((2.25*((x - 2)**2)) - 9)}

\arrow \reverse \function{-3,0,0.1}{-3 + sqrt((2.25*((x - 2)**2)) - 9)}

\arrow \function{4,7,0.1}{-3 - sqrt((2.25*((x - 2)**2)) - 9)}

\arrow \reverse \function{-3,0,0.1}{-3 - sqrt((2.25*((x - 2)**2)) - 9)}

\arrow \reverse \arrow \dashed \function{-3,7,0.1}{-1.5*x}

\arrow \reverse \arrow \dashed \function{-3,7,0.1}{(1.5*x) - 6}

\plotsymbol[3pt]{Cross}{(2,-3)}

\plotsymbol[3pt]{Asterisk}{(5.60555,-3), (-1.60555,-3)}

\tlpointsep{4pt}

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\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7}

\axislabels {y}{{$-10$} -10, {$-9$} -9, {$-8$} -8, {$-7$} -7, {$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

\normalsize

\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(y - 3)^{2}}{11} - \dfrac{(x - 1)^{2}}{10} = 1$

Center $(1, 3)$\\

Transverse axis on $x = 1$\\

Conjugate axis on $y = 3$\\

Vertices $(1, 3 + \sqrt{11}), (1, 3 - \sqrt{11})$\\

Foci $(1, 3 + \sqrt{21}), (1, 3 - \sqrt{21})$\\

Asymptotes $y = \pm \frac{\sqrt{110}}{10}(x - 1) + 3$\\

\begin{mfpic}[12][9]{-6}{8}{-4}{10}

\axes

\tlabel(8,-0.5){\scriptsize $x$}

\tlabel(0.5,10){\scriptsize $y$}

\xmarks{-5 step 1 until 7}

\ymarks{-3 step 1 until 9}

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\arrow \reverse \arrow \dashed \function{-6,8,0.1}{0.95346*(x - 1) + 3}

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\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9}

\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x + 4)^{2}}{16} - \dfrac{(y - 4)^{2}}{1} = 1$

Center $(-4, 4)$\\

Transverse axis on $y = 4$\\

Conjugate axis on $x = -4$\\

Vertices $(-8, 4), (0, 4)$\\

Foci $(-4 + \sqrt{17}, 4), (-4 - \sqrt{17}, 4)$\\

Asymptotes $y = \pm \frac{1}{4}(x +4) +4$\\

\begin{mfpic}[12][9]{-12}{4}{-1}{6}

\axes

\tlabel(4,-0.5){\scriptsize $x$}

\tlabel(0.5,6){\scriptsize $y$}

\xmarks{-11 step 1 until 3}

\ymarks{1 step 1 until 5}

\point[3pt]{(-8,4),(0,4)}

\dotted[1pt, 3pt] \polyline{(-8,5), (0,5), (0,3), (-8,3), (-8,5)}

\arrow \function{0,4,0.1}{4 + sqrt((0.0625*((x + 4)**2)) - 1)}

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\arrow \function{0,4,0.1}{4 - sqrt((0.0625*((x + 4)**2)) - 1)}

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\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x+1)^2}{9} - \dfrac{(y-3)^2}{4} = 1$

Center $(-1, 3)$\\

Transverse axis on $y=3$\\

Conjugate axis on $x=-1$\\

Vertices $(2, 3), (-4, 3)$\\

Foci $\left(-1+\sqrt{13}, 3\right), \left(-1-\sqrt{13}, 3\right)$\\

Asymptotes $y = \pm \frac{2}{3} (x+1)+3$\\

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\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(y+2)^2}{16} - \dfrac{(x-5)^2}{20} = 1$

Center $(5, -2)$\\

Transverse axis on $x=5$\\

Conjugate axis on $y=-2$\\

Vertices $(5,2), (5,-6)$\\

Foci $\left(5,4 \right), \left(5,-8\right)$\\

Asymptotes $y = \pm \frac{2\sqrt{5}}{5} (x-5)-2$\\

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\ymarks{-8 step 1 until 4}

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\axislabels {y}{{$-8$} -8,{$-7$} -7,{$-6$} -6,{$-5$} -5,{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

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\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$\dfrac{(x-4)^2}{8} - \dfrac{(y-2)^2}{18} = 1$

Center $(4, 2)$\\

Transverse axis on $y=2$\\

Conjugate axis on $x=4$\\

Vertices $\left(4+2\sqrt{2},2\right), \left(4-2\sqrt{2},2\right)$\\

Foci $\left(4+\sqrt{26},2 \right), \left(4-\sqrt{26},2\right)$\\

Asymptotes $y = \pm \frac{3}{2} (x-4)+2$\\

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\tlabel(9,-0.5){\scriptsize $x$}

\tlabel(0.5,10){\scriptsize $y$}

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\ymarks{-3 step 1 until 9}

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\axislabels {y}{ -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9}

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\end{mfpic}

\end{multicols}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{x^{2}}{3} - \dfrac{(y - 5)^{2}}{12} = 1$

Center $(0, 5)$\\

Transverse axis on $y = 5$\\

Conjugate axis on $x = 0$\\

Vertices $(\sqrt{3}, 5), (-\sqrt{3}, 5)$\\

Foci $(\sqrt{15}, 5), (-\sqrt{15}, 5)$\\

Asymptotes $y = \pm 2x + 5$

\item $\dfrac{(y + 2)^{2}}{5} - \dfrac{(x - 3)^{2}}{18} = 1$

Center $(3, -2)$\\

Transverse axis on $x = 3$\\

Conjugate axis on $y = -2$\\

Vertices $(3, -2 + \sqrt{5}), (3, -2 - \sqrt{5})$\\

Foci $(3, -2 + \sqrt{23}), (3, -2 - \sqrt{23})$\\

Asymptotes $y = \pm \frac{\sqrt{10}}{6}(x - 3) - 2$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x-3)^{2}}{25} - \dfrac{(y+1)^{2}}{9} = 1$

Center $(3, -1)$\\

Transverse axis on $y=-1$\\

Conjugate axis on $x=3$\\

Vertices $(8, -1), (-2, -1)$\\

Foci $\left(3+\sqrt{34}, -1 \right), \left(3-\sqrt{34}, -1 \right)$\\

Asymptotes $y = \pm \frac{3}{5}(x - 3) - 1$

\item $\dfrac{(y+4)^{2}}{6} - \dfrac{(x+2)^{2}}{5} = 1$

Center $(-2, -4)$\\

Transverse axis on $x=-2$\\

Conjugate axis on $y=-4$\\

Vertices $\left(-2,-4+\sqrt{6} \right), \left(-2,-4-\sqrt{6} \right)$\\

Foci $\left(-2, -4+\sqrt{11} \right), \left(-2, -4-\sqrt{11} \right)$\\

Asymptotes $y = \pm \frac{\sqrt{30}}{5}(x + 2) - 4$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(y - 7)^{2}}{16} - \dfrac{(x - 3)^{2}}{9} = 1$

\item $\dfrac{(x - 4)^{2}}{16} - \dfrac{(y - 1)^{2}}{33} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{y^{2}}{25} - \dfrac{x^{2}}{39} = 1$

\item $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 8)^{2}}{25} - \dfrac{(y - 2)^{2}}{4} = 1$

\item $\dfrac{(x - 6)^{2}}{256} - \dfrac{(y - 5)^{2}}{64} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x-1)^2 = 4(y+3)$ \\

\begin{mfpic}[15]{-4}{5}{-5}{1}

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\end{mfpic}

\vfill

\columnbreak

\item $(x-4)^2+(y+2)^2 = 9$ \\

\begin{mfpic}[20]{-1}{8}{-6}{2}

\axes

\circle{(4,-2),3}

\plotsymbol[3pt]{Cross}{(4,-2)}

\xmarks{1,4,7}

\ymarks{-5,-2,1}

\tlabel(8,-0.5){\scriptsize $x$}

\tlabel(0.5,2){\scriptsize $y$}

\tlpointsep{4pt}

\tiny

\axislabels {x}{{$1$} 1,{$4$} 4,{$7$} 7}

\axislabels {y}{{$-5$} -5, {$-2$} -2, {$1$} 1 }

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x - 2)^{2}}{4} + \dfrac{(y + 3)^{2}}{9} = 1$\\

\begin{mfpic}[15]{-1}{5}{-7}{1}

\axes

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\tlabel(0.5,1){\scriptsize $y$}

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\scriptsize

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\end{mfpic}

\vfill

\columnbreak

\item $\dfrac{(x - 2)^{2}}{4} - \dfrac{(y + 3)^{2}}{9} = 1$ \\

\begin{mfpic}[12][9]{-4}{8}{-11}{5}

\axes

\tlabel(8,-0.5){\scriptsize $x$}

\tlabel(0.5,5){\scriptsize $y$}

\xmarks{-3 step 1 until 7}

\ymarks{-10 step 1 until 4}

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\arrow \function{4,7,0.1}{-3 + sqrt((2.25*((x - 2)**2)) - 9)}

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\plotsymbol[3pt]{Asterisk}{(5.60555,-3), (-1.60555,-3)}

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\tiny

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\axislabels {y}{{$-10$} -10, {$-9$} -9, {$-8$} -8, {$-7$} -7, {$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$}{-1}, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(y + 4)^{2} = 4x$ \\

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\end{mfpic}

\vfill

\columnbreak

\item $\dfrac{(x-1)^2}{1}+\dfrac{y^2}{4}=0$ \\

The graph is the point $(1,0)$ only.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x-1)^2}{9}+\dfrac{(y+3)^2}{4} = 1$ \\

\begin{mfpic}[18]{-3}{5}{-6}{1}

\axes

\tlabel(5,-0.25){\scriptsize $x$}

\tlabel(0.25,1){\scriptsize $y$}

\xmarks{-2 step 1 until 4}

\ymarks{-5 step 1 until -1}

\ellipse{(1,-3),3,2}

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\end{mfpic}

\vfill

\columnbreak

\item $(x-3)^2+(y+2)^2=-1$ \\

There is no graph.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{(x+3)^2}{2}+\dfrac{(y-1)^2}{1} = -\dfrac{3}{4}$ \\

There is no graph.

\vfill

\columnbreak

\item $\dfrac{(y+2)^2}{16} - \dfrac{(x-5)^2}{20} = 1$ \\

\begin{mfpic}[10]{-2}{12}{-9}{5}

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\tlabel(0.5,5){\scriptsize $y$}

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\tiny

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\axislabels {y}{{$-8$} -8,{$-7$} -7,{$-6$} -6,{$-5$} -5,{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

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\end{mfpic}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item By placing Station A at $(0, -50)$ and Station B at $(0, 50)$, the two second time difference yields the hyperbola $\frac{y^{2}}{36} - \frac{x^{2}}{2464} = 1$ with foci A and B and center $(0, 0)$. Placing Station C at $(-150, -50)$ and using foci A and C gives us a center of $(-75, -50)$ and the hyperbola $\frac{(x + 75)^{2}}{225} - \frac{(y + 50)^{2}}{5400} = 1$. The point of intersection of these two hyperbolas which is closer to A than B and closer to A than C is $(-57.8444, -9.21336)$ so that is the epicenter.
\item \begin{enumerate} \setcounter{enumii}{1} \item $\dfrac{x^2}{9} - \dfrac{y^2}{27} = 1$. \end{enumerate}
\item The tower may be modeled (approximately)\footnote{The exact value underneath $(y - 330)^{2}$ is $\frac{52707600}{1541}$ in case you need more precision.} by $\frac{x^2}{12100} - \frac{(y-330)^2}{34203} = 1$. To find the height, we plug in $x = 137.5$ which yields $y \approx 191$ or $y \approx 469$. Since the top of the tower is above the narrowest point, we get the tower is approximately 469 feet tall.