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\centerline{MATH 110-02 -- Algebra Through History}
\centerline{Discussion -- Descartes and Coordinate, or ``Analytic'' Geometry}
\centerline{December 6 and 9, 2019}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
In 1637, the French philosopher and mathematician Ren\'e Descartes published 
a pamphlet called {\it La G\'eometrie} as one of a series of discussions about
particular sciences intended apparently as illustrations of his general
ideas expressed in his work {\it Discours de la m\'ethode pour bien conduire 
sa raison, et chercher la v\'erit\'e dans les sciences} (in English: Discourse
on the method of rightly directing one's reason and searching for truth in the 
sciences).  This work introduced other mathematicians to a new way of dealing
with geometry.  Via the introduction (at first only 
in a rather indirect way) of numerical coordinates to describe points
in the plane, Descartes 
showed that it was possible to \emph{define geometric objects by means
of algebraic equations} and to apply techniques from \emph{algebra} to deduce
geometric properties of those objects.  

He illustrated his new methods first by considering a problem first studied
by the ancient Greeks after the time of Euclid:
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\noindent
\emph{Given three (or four) lines in the plane, find the locus of points $P$
that satisfy the relation that the square of the distance from $P$ to the 
first line (or the product of the distances from $P$ to the first two of the 
lines) is equal to the product of the distances from $P$ to the other two
lines.}
\vskip 10pt
The resulting curves are called \emph{three-line} or \emph{four-line} loci
depending on which case we are considering.  


For instance, the locus of points satisfying the condition above for 
the four lines 
\begin{eqnarray*}
x + 2 &=& 0,\\
x - 1 &=& 0,\\
y + 1 &=& 0,\\ 
y - 1 &=& 0
\end{eqnarray*}
is shown in red in Figure 1 (on the next page).
The point shown in black is $P = (1,1)$.  Note that it satisfies the defining condition
\emph{the product of the distances from $P$ to the first two of the 
lines is equal to the product of the distances from $P$ to the other two
lines} since it lies on the line $x - 1 = 0$ so the first product is zero, 
and it lies on the line $y - 1 = 0$, so the second product is also zero.
 
The Greek mathematician Apollonius (ca. 262 BCE - ca. 190 BCE) considered
this problem in Book III of his masterwork, the {\it Conics}.  He showed via
\emph{extremely elaborate} ``synthetic'' (i.e. Euclid-style) proofs that
\emph{both the three- and four-line loci are conic sections -- the circles, 
ellipses, hyperbolas, and parabolas that are obtained as plane sections
of a cone.}  The lines in Figure 1 give a \emph{hyperbola}; 
the other conics (circles, ellipses, parabolas) can also be obtained in 
this way by varying the positions of the lines, making them meet at non-right 
angles, etc. 

\begin{figure}
\begin{centering}
\label{expformfig}
\includegraphics[width=3in,height=3in]{FourLine.png}
\caption{A four-line locus}
\end{centering}
\end{figure}

Descartes actually learned about this work by reading an account given
in the {\it Mathematical Collection} of Pappus (ca. 290 - 350 CE).  As we have
said, Pappus' 
{\it Mathematical Collection} preserved much of the earlier work of Greek 
mathematicians and earlier thinking about the complementary
roles of {\it analysis} and {\it synthesis} in mathematics.  This work
by Pappus was translated into Latin in the 16th century
and reintroduced much Greek advanced mathematics to Europe around the 
time of F. Vi\`ete, and slightly later, R. Descartes.  With his ``analytic'' 
geometry in the plane, Descartes was able to derive Apollonius's results 
in a much easier way, and he also showed how to solve analogous problems when 
the locus was described by any number $\ge 3$ of lines.  

Today, using our knowledge of coordinate geometry, we want to understand
what Descartes did, why it was such an advance, and why it essentially
(re)-united algebra and geometry.
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\noindent
{\it Questions}
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\begin{itemize}
\item[(A)]  First, what are the coordinate equations of the ``standard''
circles and ellipses if we take the center at $(0,0)$?  What is the coordinate
equation of a ``standard'' parabola?  What do ``standard'' equations of hyperbolas
look like?  You may look this up online if you need to.  Please give the URL
for the site where you found your information if you do this(!) 
Show that the equations of all these curves can be put into the form
\begin{equation}
\label{conic}
  D x^2 + E xy + F y^2 + G x + Hy + I = 0.
\end{equation}
for appropriate choices of the numerical coefficents $D,E,F,G,H,I$.
Every curve in the plane defined by an equation of this form is a conic
section (or a pair of lines, such as the curve defined by $x^2 - y^2 = (x + y)(x - y) = 0$, 
or a ``line with multiplicity 2'' such as the curve defined by $x^2 = 0$).
\item[(B)]  A line $L$ in the coordinate plane is given by the equation 
$Ax + By + C = 0$.  Let $Q = (x_0,y_0)$ be a point.  (Note:  The subscript
$0$'s just tell us that these are particular numbers rather than 
the $x$- and $y$-coordinates of a general point, which we think of 
as variables.)  The 
(perpendicular) distance from $L$ to $Q$ is given by 
$$d(L,Q) = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}$$
Using this formula, determine the distance from the point 
$$Q = \left(\frac{-1+\sqrt{5}}{2},0\right)$$
to each of the lines given above (the ones that produce the plot at the top of the 
page).  Deduce that $Q$ lies on the four-line locus for these lines.
\item[(C)] Use part B to prove Apollonius's result that starting 
from any three or four lines 
$$L_i:  A_i x + B_i y + C_i = 0, i = 1,2,3,4$$
the locus of points $P = (x,y)$ such that {\it the square of the distance
from $P$ to line $L_1$ is equal to the product of the distances from $P$
to $L_2$ and $L_3$} is defined by an algebraic equation
of total degree 2 in the coordinates of the point $P$ -- that is, 
an equation of the form \eqref{conic} from part (A)
for some real numbers $D,E,F,G,H,I$ (\emph{after possibly ignoring the absolute values}).
Do the same for the locus of points $P = (x,y)$ such that {\it the product of the
distances from $P$ to $L_1$ and to $L_2$ is equal to the product of the distances
from $P$ to $L_3$ and to $L_4$}.  
\item[(D)]  Now suppose you had \emph{six} lines $L_1,\ldots,L_6$ in the plane and you 
wanted to understand the locus of points $P = (x,y)$ that satisfy the condition
$$d(L_1,P)d(L_2,P)d(L_3,P) = d(L_4,P)d(L_5,P)d(L_6,P).$$
What would the coordinate equation of this locus look like?  (Note:  This
was far beyond anything considered by Apollonius or any of the other ancient Greeks!
With Descartes' insights, though, it's really not any more difficult than the 
cases they looked at!)
\end{itemize}
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