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\centerline{MATH 392 -- Geometry Through History}
\centerline{Discussion -- Descartes and Coordinate, or ``Analytic'' Geometry}
\centerline{March 2, 2016}
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\noindent
{\it Background}
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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 of coordinates, Descartes 
showed that it was possible to define \emph{geometric objects} by means
of \emph{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.}
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The Greek mathematician Apollonius (ca. 262 BCE - ca. 190 BCE) considered
this problem in Book III of his masterwork {\it Conics} and showed via
\emph{extremely elaborate} ``synthetic'' (i.e. Euclid-style) proofs that
\emph{both the three- and four-line problems lead to conic sections -- the
ellipses, hyperbolas, and parabolas that are obtained as plane sections
of a cone.}

For instance, the locus of points satisfying the condition above for 
the four lines $x + 2 = 0$, $x -1 = 0$, $x + 2y + 1 = 0$, and $2x - y + 2 = 0$
is shown at the top on the back of this sheet.
The point shown in black is $P = (1,4)$.  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 $2x - y + 2 = 0$, so the second product is also zero.
This case gives an ellipse, the other conics can be obtained by varying the 
positions of the lines.  

\begin{figure}
\begin{centering}
\label{expformfig}
\includegraphics[width=4in,height=4in]{FourLine.png}
\caption{A four-line locus}
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\end{figure}

Descartes actually learned about this work by reading an account given
by the later Greek mathematician Pappus (ca. 290 - 350 CE), whose work
{\it The Collection} preserved much of the earlier work of Greek 
mathematicians.  This was translated into Latin in the 16th century
and reintroduced much Greek advanced mathematics to Europe around the 
time of Descartes.  With his coordinate 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 and why it was such an advance of Apollonius's methods.
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\noindent
{\it Questions}
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\begin{itemize}
\item[(A)]  Suppose a line $L$ is given by the equation $Ax + By + C = 0$.
and $P = (\xi,\eta)$.  Show that the (perpendicular) distance from 
$L$ to $P$ is given by 
$$d(L,P) = \frac{|A\xi + B\eta + C|}{\sqrt{A^2 + B^2}}$$
(Hint:  $u = \frac{1}{\sqrt{A^2 + B^2}}(A,B)$ is a unit vector perpendicular
to the line $L$.)
\item[(B)] Use part A to prove Apollonius's result that the three- and 
four-line loci are both given by curves defined by algebraic equations
of total degree 2 in the coordinates of the point $P$ -- that is 
equations of the form:
$$ D \xi^2 + E \xi\eta + F \eta^2 + G \xi + H\eta + I = 0$$
for some real numbers $D,E,F,G,H,I$.
\item[(C)]  Explain why every equation of the form in part (B) defines 
a conic section (or perhaps a union of two lines, or a line with multiplicity
two, or a single point, or an empty set of solutions).  
\end{itemize}
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