\section{Analytic Geometry Review Module} ANalytic geometry is the description and analysis of geometric curves and shapes in terms of mathematical coordinates. The subject is concerned mainly with geometry in two or three dimensions, using various kinds of coordinate systems. The scoep of tis module is limited to two dimension, using rectangular coordinates. It will deal with a few of the most basic results concerning straight lines,circles, parabolas, and other conic sections. Thos are about all you will need initially in basic calculus and physics. \subsection{Straight Lines} First we set up our coordinate system, with n origin and the conventional $x$ and $y$ axes, with the $x$ axis horizontal and the $y$ axis perpendicular to it on the plane of the paper. (We shouldn't really call the direction of the $y$ axis ``vertical'', although we often do.) THe position of any point in the $xy$ plane is the uniquely defined by the pair of numbers $(x,y)$. TODO diagram Any straight line in the $xy$ plane can then be described by an equation of the form \[y=mx+c\] If this equation applies, we say that $y$ is a \textit{linear function} of $x$. It is equally true that $x$ is a linear function of $y$ when $m\neq 0$. When $x=0, y=c$, and when $y=0$, $x=-c/m$. These conditions define the \textit{intercepts} of the line on the $x$ and $y$ axes, respectively. If both $m$ and $c$ are positive, the general appearance of the line is as shown in the above graph. If $(x_1,y_1$ and $(x_2,y_2)$ are any two points on the line, we have \begin{align*} y_1&=mx_1+c\\ y_2&=mx_2+c \end{align*} By subtraction, we get the result: \[m=\frac{y_2-y_1}{x_2-x_1}\] This ratio, the ratio of the vertical ``rise'' to the horizontal ``run'' between any two points, is called the \textit{slope} of the line, and the equation $y=mx+c$ is called the \textit{slope-intercept} form of the equation for a line. (You might remember the slope as ``up over the over''.) A note about slopes: If we label the angle between a straight line and the positive direction of the $x$ axis as $\theta$, then we have the relation $m=\tan \theta$. This \textit{geometrical} identification of the slope of a line in terms of an angle is fine in mathematics, where $x$ and $y$ are pure numbers and the axes are assumed to be marked off with equal scale divisions. However, it is seldom necessary, and you should be warned that it loses its usefulness -- and can in fact be misleading -- when you come to graphs of the relationship between two different kinds of quantities, measured in different units. TODO diagram For example, the graph opposite might describe the motion of a car traveling at a constant speed along a straight road. On the $y$ axis we show its position $s$ measured, say, in kilometers relative to its start position; the horizontal axis shows the time $t$ measured, say, in hours. The speed of the car, in kilometers per hour, is the slope of this graph, measured as the change in $s$ divided by the change in $t$ between any two points on the line. We simply read off these values from the scales of $s$ and $t$ along the axes, and these scales are purely a matter of convenience. There is no particular meaning to the geometrical slope of the line as such. In fact, here are two graphs of a rocket's distance from the Earth until it impacts the Sun: TODO graphs Most of the graphs we draw in science, engineering, economics, etc. are constructed with scales purely for convenience. TODO graph If we have a straight line as shown, where the point $A$ $(x_0,y_0)$ is a particular point on the line and the point $P$ $(x,y)$ is any \textit{other} point, we can put \[\frac{y-y_0}{x-x_0}=m\] This can be rewritten in the form \[y-y_0=m(x-x_0\] This is the so-called \textit{point-slope} form of the equation of a line. The \textit{general} equation of a straight line may be written: \[ax+by+c=0\] You should be able to recognize and use any of the above three forms of the equation of a straight line. For most purposes, the form $y=mx+c$ is easiest and the most convenient to use, but be comfortable with all of them. \begin{exercise} Find the intercept $c$ in the slope-intercept equation in terms of the quantities $x_0$, $y_0$, and $m$ of the point-slope equation. Find the slope $m$ and the intercept $c$ of the slope-intercept equation in terms of the constants $a,b,c$ of the general equation. \end{exercise} Note: The answers to the exercises are all collected together at the end of this module. We have tried to eliminate errors, but if you find anything that you think needs to be corrected, please write to us. \begin{exercise} Draw a set of $xy$ axes, marked off in equal intervals between $-5$ and $+5$ for both $x$ and $y$, and sketch straight lines with the following values of $m$ and $c$: \begin{enumerate} \item $m=2, c=-4$ \item $m=-1,c=4$ \item $m=\frac 12, c=1$ \item $c=-1, m=-\frac 13$ \end{enumerate} \end{exercise} \begin{exercise} Draw another set of $xy$ axes, marked off as before, and draw the following lines, all passing through the point $(2,1)$: \begin{enumerate} \item $m=0$ \item $m=\frac 32$ \item $m=\infty$ \item $m=-\frac 23$ \end{enumerate} \end{exercise} \begin{exercise} Draw yet another set of $xy$ axes, marked off as before, and draw lines described by the following equations: \begin{enumerate} \item $x+y=2$ \item $2x-y+3=0$ \item $x+2y+4=0$ \item $x-4=0$ \end{enumerate} \end{exercise} By definition, \textit{lines with the same value of $m$ have the same slope. Lines with the same value of $m$ but different values of $c$ are therefore parallel.} If two lines are perpindicular, the product of their slopes is 1. We included an example of this in Exercise \ref{ana:ex:three:one:three}. The result is easily shown: here is one way. Let the point of intersection of two perpindicular lines be called $O'$. Take $O'$ as a new origin, and mark off a distance $l$ along each line as shown in the diagram. If $\angle AO'M=\alpha$, then $\angle O'BN$ is also $\alpha$. With respect to $O'$, the coordinates of $A$ are $(x_1,y_1)$, where $x_1=l\cos \alpha$, $y_1=l\sin\alpha$, and we then have \[m_1=\frac{y_1}{x_1}=\tan\alpha\] Also, with respect to $O'$, the coordinates of $B$ are $(x_2,y_2)$, where $x_2=-l;\sin\alpha$,y$y_2=l;\cos\alpha$, and so we have:\[m_2=\frac{y_12}{x_2}=\frac {l\cos\alpha}{-l\sin\alpha}=-\cot\alpha\] Therefore, $m_1m_2=\left(\tan\alpha\right)\left(-\cot\alpha\right)=-1$ Unless two planes in the $xy$ plane are parallel, they must intersect somewhere. To find the point of intersection, we use the condition that this point must be onboth lines, which gives us a pair of simultaneous equations. For instance, find the point of intersection of the lines $2x-y-2=0$, $x-2y+2-0$. Let the point of intersection be $(x_1,y_1)$. Then: \begin{align*} 2x_1=y_1&=2\\x_1-2y_1&=-2 \end{align*} Solving these equations gives $x_1=2$< $y_1=2$. (Check this!) \textit{Drawing sketches for such problems is highly recommended!} \begin{exercise} Find the point of intersection of the lines $3x-4y-6=0-$ and $x+2y+3=0-$.\end{exercise} \begin{exercise} Which of the following lines are parallel? perpendicular? intersect? \begin{enumerate} \item $y=2x+4$ \item $2y+x-1=0$ \item $y=\frac{-3}{2}x+1$ \item $3x+2y=-1$ \item $2=(y-4)/x$ \end{enumerate} \end{exercise} \subsection{Conic Sections} You have probably learned that the circle, the parabola, the ellipse, and the hyperbola are called ``conic sections.'' This is because they arise from the intesection of various planes with a circular cone. The diagram below shos how this comes about. Circles are formed when a plane is drawn at right angles to the axis of the cone. Parabolas are formed when a plane is drawn parallel to one of the generators (sloping sides) of the cone. Ellipses are formed by planes intermediate between these two directions. Hyperboolas are formed by planes that intersect both top and bottom parts of the copmlete double cone: the diagram shows the articular plane that is parallel to the axis of the cone. TODO diagram From Hughes-Hallett, \textit{Math Workshop: Elementary Functions}, W. W. Norton Co. (1980). We will not make sue of these pictures to obtain equations for conics: that would be partciularly ugly. Instaead, we',ll work from geometric definitions of how these curves are constructed in terms of a rectangular coordinate system. But it is worth seeing how they are members of a family. subsubsection{Circles} There are just two features that characterize a circle: it has a certain radius $r$ that is constant; and its center is at a certain point $(x_0,y_0)$. To expressa these properties in mathematical terms, we combine them in a statement that the distacne between the center of the circle and any pouitn $(x,y)$ on its circumference is equal to $r$. For any two points on the $xy$ plane, the square of the distance between them is given by the Pythagorean Theorem. The equation of the circle is therefore given by:\[(x-x_0)^2+(y=y_0)^2=r^2\] This is the \textit{standard form} of the eqyuation of a circle. If the center of the circle is at the origin, this reduces to the simple form: \[x^2+y^2=r^2\] \begin{exercise} Write the equations of the following circles, and sketch them on graph paper: \begin{enumerate} \item Center at (3,2), radius 3; \item Center at (0,3), radius 5; \item Center at (-2,2), passing through (2,5) \end{enumerate} \end{exercise} If the equation of a circle with its center at an earbitrary point is expanded out into individual terms and the contributions to the constant term are collected together, the resulting equation is of the form \[x^2+y^2+ax+by+c=0\] It is not necessarily true, however, that any equation of the abocve form descibes a circle. We can see this by completing the square for both terms. If we do this, and rearrange, we get: \[\left()x+\frac{a}{2}\right)^2+\left(y+\frac b2\right)^2=\frac{a^2}{4}+\frac{b^2}{4}-c\] The right-hand side can be positive, negative, or zero. If it is positive, we can equate it to $r^2$, the equation does describe a circle -- with its center at $(-\frac a2,-\frac b2)$ If it is negative, there are no real values of $ x$ and $y$ that satisfy the equation. The sum of two squares must be greater than 0, since any one square is greater than 0. The equation cannot describe any real curve or point. If it is zero, the equation can be satisfied only for a sinhgle point: $x=-a/2$, $y--b/2$, since both of the squares on the left myust be xero if the equation is to be satisfied. This is, if you like, a circle of zero radius, againh with its center at $(-a/2,-b/2)$. \begin{exercise} Analyze which of these equations describ circles. For those that do, find the radius and the coordinates of the center: \begin{enumerate} \item $x^2+y^2-2x+89=0$ \item $x^2+y^2-4x+2y+6=0$ \item $x^2+y^2+4x-6y-12=0$ \item $x^2+y^2-8x+6y+26=0$ \end{enumerate} \end{exercise} \begin{exercise} Here is a nice exercise that puts together the analytic geometry of straight lines and circles. Take a circle of radius $r$ with its center at the origin. See if you can prove, by analytic geometry, the theorem from plane geometry that ssays that an angle inscribed ina semicircle is a right angle. [First find the slopes of the lines $AP$ and $BP$ in terms of $x$, $y$, and $r$. To confirm the perpendicularity condition $m_1m_2=-1$; use $x^2+y^2=r^2$.] TODO diagram \end{exercise} \subsubsection{Parabolas} A parabola is defined geometrically as a plane curve that is the locus of points equidistant from a fixed point (the \textit{focus},l $F$) and a fixed straight line (the \textit{directrix}). The word focus can be taken literally: a mirror made into a parabolic (or, more strictly, a paraboloidal) shape will bribring parallel light to a focus at $RF$. (A paraboloid is a parabola rotated about its axis of symmetry.) Suppose that the foucs is at the point $(p,0()$, and that the directrix is the line $x=-p$. Youc an see right away that the oribgin is a point on the curve; it is halfway between the focus and the directix. For any other point $P=(x,y)$, we have the more complicated condition that the distacne $FP$ is equal to the length of the perpendicular $PN$ drawn from $P$ to the directrix. Using the Pythagroean Theeam, this requires \[(FM)^2+(MP)^2=(PN)^2\] Putting $FM=x-p$, $MP=y$, and $PN=x+p$, this gives: \[(x-p)^2+y^2=(x+p)^2\] Multiplying this out, we get \[x^2-2px+p^2+y^2=x^2+2px+p^2\] which after canceling and rearraanging gives \[hy^2=4px\] This equation describes a parabola with its axis along the $x$ axis and its \textit{vertex} (the rounded end) at the orign. The parabola is symmetrical about the $x$ axis, because for any given value of $x$ we have $y=\pm\sqrt{4px}$. The coefficient $4p$ has a geometrical significance. If we draw the line $x=p$ through the focus, it uts the parabola at points $A$ and $B$ for which $ty^2=4p^2$ and so $y=+\pm 2p$. Thus the distance between these points is equal to $4p$; it is a measure of of the width of the parabola. We can construct similar parabolas with different positions of the focus and different locations and directions of the directrix. For example, a aparabola with the same value of $p$ as before (i.e., the same fdistance between focus and directrix), with its vertex at $(h,hk0$ and its directrix parallel to the $y$ axis has the equation \[(y-k)^2=4p(x-h)\] TODO diagram The figures below show some other examples. TODO figures of $x^2=4py$, $x^2=0-4py$, and $(x-h)^2=4p(y-k)$ The first of these, whose equation can be rewritten in the form $y=Cx^2$, iis a particularly useful one to rememvber, because we often encounter relationships between two quantities where the dependent variable $y$ is porportional to the square of the independent variable ($x$). For instance, the distance $y$ as a function of time $x$ that an object falls from rest under gravity. \begin{exercise} Take some graph paper and sketch the following parabolas: $y=-x^2$ $x=2(y-1)^2$ $y=(x+2)^2$ \end{exercise} There are two standard forms for teh equation of a parabola,. depending on whether its axis is along the $x$ direction or the $y$ direction. Notice that if the axis is parallel to $x$, the equation is quadratic in $y$; if the axis is parallel to $y$k, the equation is quadratic in $x$. You can of course have parabolas whose axis is in some arbitrary direction, but we won't bother with those . Try to figure out what their equations would look like, though! If the axis is parallel to $x$, the equation can be written in the form \[x^2+ax+by+c=-\] As with the circle, we can put this into standard form, beginning by completing the square on the left side: \[\left(x+\frac a2\right)^2=-by-c+\frac{a^2}{4}=-b\left(y+\left(\frac cb-\frac{a^2}{4b}\right)\right)\] Comparing this with $(x-h)^2=4p(y-k)$, we can find the values of $h$, $k$, and $p$. For example, we can find the coordinates of the focus and the vertex of the parabola $x^2-6x-8y+17-0$. Completing the square on the $x$ terms and rearranging, we have $(x-3)^2=8y-17+9=8y-8=8(y-1)$. Thus the vertex is at $(3,1)$, and $p=2$. In this parabola, the focus ais a distance $p$ above the vertex; its coordinates are therefore (3,3). \begin{exercise} The following equations are in the general polynomial form of tequations for parabolas. Put them into standard form, locate the vertices and foci, and sketch the curves: \begin{enumerate} \item $x^2-4x-4y=0$ \item $12x-6y-y^2-33=0$ \item $y^2+4y+4x=8$ \item $2x^2-4x-y+1=0$ \end{enumerate} \end{exercise} \subsubsection{Ellipses} TODO diagram of pin/string You are very likely familiar with the ``pins-and-string'' definition of an ellipse. Take a piece of string and attach its ends to pins at two fixed points, $F$ and $F'$, on a piece of paper. Take a sharp pencil and hold it vertically. Move it until it just makes the string tight, and then move it around, always in contact with the string, so as to create a closed curve. This curve is an ellipse. If the length of the string is called $2a$, we have:\[FP+F'P=2a\] The points $F$ and $F'$ are the \textit{foci}. If the ellipse were a mirror, light from a point source at $F$ would come to a focus at $F'$, and vice versa. The ration of the distance $OF$ to the distance $OA$ is called the \textit{eccentricity}, $\epsilon$ of the ellipse: \[\epsilon=\frac{OF}{OA}\] (Note that, if $F$ and $F'$ are both moved to the center $O$, the ellipse becomes a circle and by the above definition the eccentricity is zero.) Take an origin at the geometrical center of the figure, with an $x$ axis along the line $F'F$ ($AA'$), and a $y$ axis along $OB$. Let the distance $OF$ (${}=OF'$) be $L$, and let $OB$ be $b$. TODO diagram repeated. First, imagine that $P$ is moved to $A$. Then $FA+AF'=2a$> But, by symmetry, $FA=F'A'$. Therefore $AF'+F'A'$ (which adds up to $AA'$) is equal to $2a$. So $OA=a$. Next, imagine $P$ is moved to $B$; this gives $FB$ (${}=BF'$) ${}=a$. Using Pythagoras' Theorem, one finds $b=\sqrt{a^2-l^2}$. Finally, let $P$ be at some arbitrary point $(x,y)$ on the ellipse. From the condition $FP+PF'=2a$, and using Pythagoras' Theorem again, we have:\[\sqrt{(l-x)^2+y^2}+\sqrt{(l-x)^2+y^2}=2a\] If you do some algebra on this, you can verify that it leads to the following equation for the ellipse in $xy$ coordinates: \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] The distances $a$ and $b$ are called the \textit{semi-major axis} and the \textit{semi-minor axis}, respectively ($a>b$). \begin{exercise} Go through the process of deriving this last equation from the preceding one. [Begin by putting one of the radicals on the other side of the equation and squaring. Some rearrangement, followed by another squaring, will do most of the rest, but you will need the relation between $a$, $b$, and $l$.] \end{exercise} If the center of the ellipse is not at the origin but at some point $(h,k)$, one just replaces $x$ by $(x-h)$ and $y$ by $(y-k)$ in the equation. The resulting equation, when multiplied out, will in general have quadratic and linear terms in $x$ and $y$, plus a constant. The thing that immediately distinguishes the equation of an elliose from the equation of a circle is that, for a circle, the coefficients of the $x^2$ and $y^2$ terms are equal, whereas for an ellipse they are different. \begin{exercise} Sketch the ellipse $4x^2+9y^2=36$, locate its foci and find the eccentricity. \end{exercise} For $b=a$ the equation of an ellipse turns into the equation for a circle, and it is clear that the connection between the two is close. In fact, an ellipse can be though of as the geometrical projection of a circle onto another plane. The diagram here indicates this. If the angle between the planes is $\theta$, then a circle of radius $b$ projects into an ellipse of semi-minor axis $b$ and semi-major axis $a=b\sec\theta$. If the coordinates of a point $P$ on the circle are labeled $(x,y)$, the coordinates of the corresponding point $P'$ on the ellipse are $(x\sec\theta,y)$. Thus every $x$-coordinate on the ellipse is stretched by the factor $\sec \theta = a/b$. One can use this fact to infer that the \textit{area} of an ellipse is given by $\boxed{\pi ab}$. TODO diagram of projection. \subsubsection{Hyperbolas} TODO generic hyperbola x-axis. The way of constructing a hyperbola looks very similar to that for an ellipse, but is not nearly so easy to do in practice. Again one has an axis with two foci on it. However, the prescription for a hyperbola is to move a point $P$ so that the \textit{difference} of its distances from the foci is constant. If the foci are the $x$ axis, the $xy$ equation is \[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\] This is rather like an ellipse turned inside out. The distance intercepted by the curve on the $x$ axis is again $2a$, but this time it is a minimum distance rather than a maximum. At large distances from the origin, the two branches of the curve approach two straight lines --- the \textit{asymptotes} --- whose equations are the two possibilities allowed by the equation $(x^2/a^2)-(y^2/b^2)=0$:\[y=\pm\frac ba x\] TODO generic y-axis hyperbola. If the foci lie on the $y$ axis, the branches of the hyperbola are as shown in the second diagram. The equation to the curve is \[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\] and the asymptotes are given by \[y=\pm \frac abx\] In other words, the whole picture is turned through $90^\circ$ with respect to the first one. \begin{exercise} Using the condition $FP-F'P=2a$, together with the standard equation of a hyperbola, show that the distance between the foci is $2\sqrt{a^2+b^2}$. \end{exercise} ANother very important form of the equation of a hyperbola --- and in some respects much the simplest, is \[xy=c\] where $c$ is a constant. TODO asymptote=axis hyperbolas. This gives the two possibilities shown here. This is the \textit{rectangula hyperbola} --- its asymptotes are perpindicular. A good physical example of this (with $c>0$) is the relation between pressure and volume for an ideal gas at constant temperature: \[pV=\text{constant}\] \begin{exercise} Sketch the hyperbola $xy=9$, and locate its foci. \end{exercise} \section{Answers to Exercises} \textit{Answers not given to exercises \ref{ana:ex:five:one}, \ref{ana:ex:six:one}} Exercise \ref{ana:ex:one:one}: (a) $c=y_0-mx_0$; (b) It is very confusing to use the same letter ($c$) for two different quantities in a problem. Before solving, we change the name of $c$ to $d$ in the general equation, $ax+by+d=0$, and leave it as $c$ in the slope-intercept equation, $y=mx+c$. Then $m=-a/b$, $c=-d/b$. Exercise \ref{ana:ex:one:two}: TODO sketch Exercise \ref{ana:ex:one:three}: TODO sketch Exercise \ref{ana:ex:one:four}: TODO sketch Exercise \ref{ana:ex:one:five}: $x=-1/5, y=-7/5$. Exercise \ref{ana:ex:one:six}: $\ell_c\parallel \ell_d$, $\ell_b\perp \ell_a$, $\ell_b\perp \ell_e$. ($\ell_e,\ell_a$ are the same line). All non-parallel lines intersect! Exercise \ref{ana:ex:three:one}: (a) $(x-3)^2+(y-2)^2=9$ TODO diag; (b) $x^2+(y-3)^2=25$ TODO diag; (c) $(x+2)^2+(y-2)^2=25$ TODO diag Exercise \ref{ana:ex:three:two}: (a) No; (b) No; (c) Center at $(-2,3)$, $r=5$; (d) Single point, $(4,-3)$. Exercise \ref{ana:ex:three:three}: \begin{align} m_1 &=\text{slope }AP=\frac{y-0}{x-r}\\ m_2 &=\text{slope }BP=\frac{y-0}{x-(-r)}\\ m_1m_2 &=\frac{y}{x-r}\cdot \frac{y}{x+r}=\frac{y^2}{x^2-r^2} \end{align} But $x^2+y^2=r^2$ so $x^2-r^2=-y^2$ so $m_1m_2=y^2/(-y^2)=-1$. Exercise \ref{ana:ex:four:one}: (a) TODO diag; (b) TODO diag; (c) TODO diag Exercise \ref{ana:ex:four:two}: (a) $(x-2)^2=4(y+1)$; (b) $(y+3)^2=12(x-2)$; (c) $(y+2)^2=-4(x-3)$; (d) $(x-1)^2=\frac 12 (y+1)$ TODO diag x4 Exercise \ref{ana:ex:five:two}: $(\sqrt{5},0),(-\sqrt{5},0)$; $\epsilon=\ell/a=\frac{\sqrt{5}}3$ TODO diag Exercise \ref{ana:ex:six:two}: Foci at $(\sqrt{13},0),(-\sqrt{13},0)$ TODO diag Exercise \ref{ana:ex:six:three}: Foci at $(3\sqrt{2},3\sqrt{2}),(-3\sqrt{2},-3\sqrt{2})$ TODO diag \emph{This module is based largely on an earlier module prepared by the MIT Mathematics Department.} \section{Review Problems for Analytic Geometry} \subsection{Lines} \begin{problem} A line of slope 6 passes through the point $(2,-3)$. Where does it cross the $x$-axis? \end{problem} \begin{problem} At what point do the lines given by $3x+4y=12$ and $3x-5y=-6$ intersect? \end{problem} \begin{problem} At what point does a line with slope $-1$ passing through the point $(2,5)$ intersect the line having slope 2 and $y$-intercept $-2$? \end{problem} \begin{problem} What is the equation of a line perpindicular to the line $2x-5y=1$ and having $y$-intercept 3? \end{problem} \begin{problem} The rectangle $ABCD$ is symmetric about the $y$-axis. What is the equation of a line through $C$ parallel to the diagonal $BD$? $B=(5,3), C=(5,-1)$. TODO diagram \end{problem} \begin{problem} Find the area of the finite region bounded by the lines \[x=0;\hspace{.5in}y=0,\hspace{.5in} 3x+5y=30\] \end{problem} \begin{problem} Find the equation of a line through the origin perpindicular to the line having $y$-intercept $-2$ and $x$-intercept 3. (Draw a sketch.) \end{problem} \subsection{Circles} \begin{problem} Give in the form $x^2+y^2+ax+by+c=0$ the equation of the circle \begin{enumerate} \item with center at the origin and radius 3 \item with center at the origin and tangent to the line $x+y=1$. \item with center at $(1,1)$ and passing through the origin \item with center at $(-1,1)$ and tangent to both coordinate axes. \end{enumerate} \end{problem} \begin{problem} A circle centered at the origin has radius 11. Is the point $(7,8)$ outside or inside the circle? \end{problem} \begin{problem} Where do the circle and line intersect in the following examples? \begin{enumerate} \item $x^2+y^2=5$, $x+y=3$ \item $x^2+y^2=10$, $x+3y=10$ \item circle of radius 3 centered at $(0,0)$, line of slope 3 through $(0,0)$ \end{enumerate} \end{problem} \subsection{Conics} \begin{problem} What is the length of the semi-major and semi-minor axes of the ellipses \[\frac{x^2}{10}+\frac{y^2}3=1\]\[9x^2+4y^2=25\] \end{problem} \begin{problem} Write the equation of the ellipse with center at the origin, axes along the two coordinate axes, and whose $x$-intercepts are $\pm 3$, $y$-intercepts are $\pm 2$. \end{problem} \begin{problem} Write the equation of a parabola whose minimum point is on the $y$-axis at $-1$, and whose $x$-intercepts are $\pm 2$. \end{problem} \begin{problem} Write in the form $y=ax(x-c)$ the equation of a parabola whose high point is at $(2,1)$, and which goes through the origin. \end{problem} \begin{problem} Write the general form for the equation of a hyperbola having the lines $y=\pm 2x$ as asymptotes. \end{problem} \begin{problem} What kind of geometric locus do the following equations represent: circle, ellipse, parabola, hyperbola, straight lines, point, or none? \begin{enumerate} \item $x^2+2y^2-12=0$ \item $6x^2-5y^2=0$ \item $15-x^2-y^2=0$ \item $x^2+2y-3=0$ \item $xy=2$ \item $4y^2-x^2=2$ \item $x^2+y^2+2x-5=0$ \item $y^2+2x=5$ \item $x^2-y^2=0$ \item $x^2+2y^2+1=0$ \item $xy=5$ \item $x^2+y=0$ \item $x^2-y^2=4$ \item $2x^2+3y^2=5$ \item $xy=0$ \item $3x^2+3y^2-6y=10$ \item $x^2-5y^2=10$ \item $x^2+3y^2=0$ \item $x+3y^2-5=0$ \item $x^2-4y^2=0$ \end{enumerate} \end{problem} \section{Solutions to Analytic Geometry Review Problems} \subsection{Lines} \begin{solution} Pt.-slope equation of line is $y+3=6(x-2)$. Crosses $x$-axis when $y=0$, i.e. $6(x-2)=3$, so $x-2=\frac 12$, $x=\boxed{\frac 52}$ \end{solution} \begin{solution} Solve simultaneously $3x+4y=12$ and $3x-5y=-6$. Subtracting, $9y=18$, so $y=2, x=\frac 43$; ans: $\boxed{(\frac 43,2)}$ \end{solution} \begin{solution} Slope $-1$, through $(2,5)$: $y-5=-(x-2)$ or $y=-x+7$. Slope 2, $y$-intercept $-2$: $y=2x-2$ Solving simultaneously, $2x-2=-x+7\rightarrow x=3,y=4$. Ans: $\boxed{(3,4)}$ \end{solution} \begin{solution} $2x-5y=1\rightarrow y=\frac 25 x -\frac 15$, slope is $\frac 25$. Thus, slope of perpindicular line is $-\frac 52$ (negative reciprocal). Since $y$-intercept is $3$, its equation is $\boxed{y=-\frac 52 x +3}$. \end{solution} \begin{solution} TODO diagram slope of $BD$ is $\frac{3-(-1)}{5-(-5)}=\frac{4}{10}=\frac 25$; thus $L$ has equation $\boxed{y+1=\frac 25(x-5)}$ (or $5y-2x+15=0$). \end{solution} \begin{solution} TODO diagram The $x$-intercept is where $y=0$, so $3x=30$, $x=10$ $y$-intercept is where $x=0$, so $5y=30$, $y=6$. Thus the area of the triangle is $\frac 12 \cdot 6\cdot 10=\boxed{30}$. \end{solution} \begin{solution} TODO diagram $L$ has slope $\frac 23$ so $L'$ has slope $-\frac 32$. $\boxed{y=-\frac 32 x}$. \end{solution} \subsection{Circles} \begin{solution} TODO 3 diagrams \begin{enumerate} \item $x^2+y^2-9=0$ \item $x^2+y^2-\frac 12=0$ \item $(x-1)^2+(y-1)^2=2\rightarrow x^2+y^2-2x-2y=0$ \item $(x+1)^2+(y-1)^2=1^2\rightarrow x^2+y^2+2x-2y+1=0$ \end{enumerate} \end{solution} \begin{solution} The equation is $x^2+y^2=11^2$. \textit{Inside} is where $x^2+y^2\leq 11^2$, \textit{outside} is where $x^2+y^2>11^2$. Here $7^2+8^2=113<11^2=121$. Thus $(7,8)$ is \textit{inside} the circle. \end{solution} \begin{solution} \begin{enumerate} \item $x^2+y^2=5$, $x+y=3$. Solving simultaneously, $y=3-x$; substitute into eqn of circle: $x^2+(3-x)^2=6\rightarrow 2x^2-6x+4=0$, or $x^2-3x+2=0\rightarrow (x-2)(x-1)=0$. Thus, the solutions are $x=2,y=1$ and $x=1,y=2$. \item $x^2+y^2=10$, $x+3y=10$ gives $(10-3y)^2+y^2=10\rightarrow 100-60y+10y^2=10\rightarrow y^2-6y+9=0\rightarrow (y-3)^2=0$. Thus, $x=1,y=3$ is the solution. (Note that there is only one solution because the line is \textit{tangent} to the circle.) \item $x^2+y^2=9$, $y=3x$ gives $x^2+9x^2=9\rightarrow 10x^2=9\rightarrow x=\pm\frac{3}{\sqrt{10}},y=\pm\frac{9}{\sqrt{10}}$. Answers: $(\frac{3}{\sqrt{10}},\frac{9}{\sqrt{10}}), (-\frac{3}{\sqrt{10}},-\frac{9}{\sqrt{10}})$. \end{enumerate} \end{solution} \subsection{Conics} \begin{solution} a: Intercepts are when $x=0$, $y=\pm\sqrt{3}$ and $y=0$, $x=\pm \sqrt{10}$. Thus, semimajor axis is $\sqrt{10}$, semiminor axis $\sqrt{3}$. TODO diag b: Intercepts are at $x=0$, $y=\pm\frac 52$ and $y=0$, $x=\pm\frac 53$. Thus, semimajor axis is $\frac 52$, semiminor axis is $\frac 53$. TODO diag \end{solution} \begin{solution} TODO diag \[\left(\frac x3\right)^2+\left(\frac y2\right)^2=1\text{ or } \frac{x^2}9+\frac{y^2}4=1\] \end{solution} \begin{solution} TODO diag Equation is $y=ax^2-1$ (parabola $y=ax^2$ moved down by one unit). Choose $a$ so that it goes through $(2,0)$: $0=a\cdot 4-1\rightarrow a=\frac 14$. Thus, we have $\boxed{y=\frac{x^2}4-1}$. [Another way: since roots are at $\pm 2$, equation is $y=c(x+2)(x-2)=c(x^2-4)$, and choose $c$ so that $(0,-1)$ is on parabola.] \end{solution} \begin{solution} TODO diag Since high point is at $x=2$ and it goes through origin, other intercept is $x=4$. (High point is midway between the two $x$-intercepts.) Thus, $y=ax(x-4)$; choose $a$ so that $1=a\cdot 2(2-4)\rightarrow a=-\frac 14$; $\boxed{y=-\frac 14 x(x-4)}$ is our final answer. \end{solution} \begin{solution} The lines are $y+2x=0$ and $y-2x=0$; taken together, their equation is $(y+2x)(y-2x)=0$ or $y^2-4x^2=0$. Eqn of hyperbola is then $y^2-4x^2=c$ ($c$ can be positive or negative but not 0). \end{solution} \begin{solution} \begin{enumerate} \item ellipse \item two lines \item circle \item parabola \item hyperbola \item hyperbola \item circle \item parabola \item two lines \item none \item hyperbola \item parabola \item hyperbola \item ellipse \item two lines \item circle \item hyperbola \item point $(0,0)$ \item parabola \item two lines \end{enumerate} \end{solution} \section{Analytic Geometry Diagnostic Test} \begin{problem} One line has a slope of $-1$ and passes through the point $(2,5)$. Another line is given by the equation $2x-y-2=0$. At which point do they intersect? \end{problem} \begin{problem} A line with a slope of $\sqrt{3}$ passes through the point $(2,-3)$. Where does it cross the $x$-axis? \end{problem} \begin{problem} Circle $A$ is defined by $x^2+y^2=2$, and line $B$ is defined by $x+y-1=0$. Find the points where $A$ intersects $B$. \end{problem} \begin{problem} What kind of geometrical locus do the following equations represent: circle, ellipse, parabola, hyperbola, point, or none? \[2y-4x^2=3\] \[x^2+3y^2=6\] \end{problem} \section{Analytic Geometry Diagnostic Test Solutions} \begin{sproblem} One line has a slope of $-1$ and passes through the point $(2,5)$. Another line is given by the equation $2x-y-2=0$. At which point do they intersect? \end{sproblem} \begin{solution} First line: $y-5=-(x-2)$ or $x+y=7$; now solve $x+y=7$ and $2x-y=2$, to obtain $x=3, y=4$. Answer: $\boxed{(3,4)}$. \end{solution} \begin{sproblem} A line with a slope of $\sqrt{3}$ passes through the point $(2,-3)$. Where does it cross the $x$-axis? \end{sproblem} \begin{solution} Line is $y+3=\sqrt(3)(x-2)$ Crosses $x$-axis where $y=0$; solving for $x$, we have $\sqrt{3}(x-2)=3\rightarrow x-2=\sqrt{3}\rightarrow \boxed{x=2+\sqrt{3}}$ \end{solution} \begin{sproblem} Circle $A$ is defined by $x^2+y^2=2$, and line $B$ is defined by $x+y-1=0$. Find the points where $A$ intersects $B$. \end{sproblem} \begin{solution} Solve simultaneously: $x^2+y^2=2$ and $x+y-1=0$. We eliminate $y$ to get $y=1-x$ from the second equation, and substitute: $x^2+(1-x)^2=2\rightarrow 2x^2-2x-1=0$. This has solutions $x=\frac{2\pm\sqrt{4+8}}4=\frac{1\pm\sqrt{3}}2$, $y=\frac{1\mp\sqrt{3}}2$; so the intersection points are $\boxed{(\frac{1+\sqrt{3}}2,\frac{1-\sqrt{3}}2),(\frac{1-\sqrt{3}}2,\frac{1+\sqrt{3}}2)}$. \end{solution} \begin{sproblem} What kind of geometrical locus do the following equations represent: circle, ellipse, parabola, hyperbola, point, or none? \[2y-4x^2=3\] \[x^2+3y^2=6\] \end{sproblem} \begin{solution} Parabola, ellipse. \end{solution}