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{\bf Coordinate Geometry}
\end{center}

{\bf Conic sections}

These are plane curves which can be described as the intersection
of a cone with planes oriented in various directions.

It can be demonstrated that the locus of a point which moves so
that its distance from a fixed point (the focus) is a constant
multiple ($e$ - the eccentricity) of its distance from a fixed
straight line (the directrix) is a conic section.

If $e&lt;1$ we obtain an ellipse.

If $e=1$ we obtain a parabola.

If $e&gt;1$ we obtain a hyperbola.

See Scientific American September 1977-Mathematical games section
p24.

${}$

{\bf Cartesian equation}

Take as the x-axis a line perpendicular to the directrix passing
through the focus.  Take the origin to be where the conic cuts the
axis between the focus and directrix.


DIAGRAM


From the definition of a conic $\ds SP^2=e^2PM^2$

$\ds y^2+(x-ek)^2=e^2(x+k)^2$

$\ds y^2+x^2-2ekx+e^2k^2=e^2x^2+2e^2kx+e^2k^2$

$\ds y^2+x^2\left(1-e^2\right)-2ke(1+e)x=0$

If we have a parabola where $e=1$ then the equation reduces to
$\ds y^2=4kx$.

If $e\not=1$ we write the equation in the form

$\ds
\frac{y^2}{1-e^2}+\left(x-\frac{ke}{1-e}\right)^2=\frac{k^2e^2}{\left(1-e\right)^2}$

We now write $\ds \frac{ke}{1-e}=a$, and shift the origin to the
point $(a,0)$.  Referred to these new axes the equation becomes

$\ds\frac{x^2}{a^2}+\frac{y^2}{a^2\left(1-e^2\right)}=1$

The focus becomes the point $(-ae,0)$ and the directrix the line
$\ds x=-\frac{a}{e}$.

Notice that the equation is unchanged if $x$ is replaced be $-x$,
so that there is a second focus at $x=(ae,0)$ and a second
directrix at $x=\frac{a}{e}$.

For an ellipse $e&lt;1$ and we write $b^2=a^2(1-e^2)$ so the equation
becomes

$\ds \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

For a hyperbola $e&gt;1$ and we write $\ds b^2=a^2(e^2-1)$ so the
equation becomes

$\ds \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

{\bf Focal distance properties}

Ellipse $(e&lt;1)$


DIAGRAM


From the definition

\begin{eqnarray*} S_1P+S_2P &amp;=&amp; ePM_1+ePM_2=e(PM_1+PM_2)\\
&amp;=&amp;eM_1M_2=e\frac{2a}{e}=2a\end{eqnarray*}

So the sum of the focal distances is constant.

Hyperbola $(e&gt;1)$


DIAGRAM


From the definition

\begin{eqnarray*} S_2P-S_1P&amp;=&amp;ePM_2-ePM_1=e(PM_2-PM_1)\\
&amp;=&amp;e\frac{2a}{e}=2a\end{eqnarray*}

Similarly

$\ds S_1Q-S_2Q=2a$

{\bf The Parabolic Mirror}


DIAGRAM


Suppose a ray of light comes in parallel to the x-axis and is
reflected in a direction equally inclined to the tangent.  We
prove that it passes through the focus.

Let the parabola have equation

$\ds y^2=4kx$, so $\ds S=(k,0), \,\,\,\, P=(x,y)$

$\ds 2y\frac{dy}{dx}=4k$ so $\ds \frac{dy}{dx}=\frac{2k}{y}$

thus $\ds \tan\alpha_1=\frac{2k}{y}$.

Now $\ds \tan\alpha_3=\frac{y}{x-k}$ and $\ds
\alpha_2=\alpha_3-\alpha_1$

So $\ds
\tan\alpha_2=\tan(\alpha_3-\alpha_1)=\frac{\tan\alpha_3-\tan\alpha_1}
{1+\tan\alpha_3\tan\alpha_1}=\frac{2k}{y}$ (verify)

so $\ds \alpha_1=\alpha_2$

So a parallel beam of light will be reflected through the focus.


{\bf Parametric equations}

Because a curve is one-dimensional we can label the points by
means of a single real variable, as in the following examples.
Traditionally the letter $t$ is used as the parameter, analogous
with the curve being traced out in time.

Examples

\begin{itemize}
\item[i)]
$x=a+t, \,\,\,\, y=b+mt$ represents the straight line through
$(a,b)$ with slope $m$.

\item[ii)]
$x=a\cos t, \,\,\,\, y=a\sin t$ represents the circle of radius a
centred at (0,0).  We use $\ds \cos^2+\sin^2=1, \,\,\,\, t$
corresponds to an angle and so $\theta$ is sometimes used.

\item[iii)]
$x=a\cos t, \,\,\,\, y=b\sin t$ represents the ellipse $\ds
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ again $t$ represents an angle
but not the angle from $O$ to $P$.


DIAGRAM


\item[iv)]
to parameterise a hyperbola we need to find $\ds \frac{x}{a}=f(t),
\,\,\,\, \frac{y}{b}=g(t)$ so that $\ds f(t)^2-g(t)^2=1.$

There are several possibilities

\begin{itemize}
\item[a)]
$\ds \frac{x}{a}=\frac{1}{2}\left(t+\frac{1}{t}\right) \,\,\,\,
\frac{y}{b}=\frac{1}{2}\left(t-\frac{1}{t}\right)$

\item[b)]
$\ds x=a\sec t \hspace{0.5in} y=b\tan t$

\item[c)]
$\left. \begin{array}{l} \ds
\frac{x}{a}=\frac{1}{2}\left(e^t+e^{-t}\right)=\cosh t\\ \ds
\frac{y}{b}=\frac{1}{2}\left(e^t-e^{-t}\right)=\sinh t
\end{array}\right\} \begin{array}{l} {\rm These\ are\ called}\\
{\rm hyperbolic\ functions.}\end{array}$
\end{itemize}

\item[v)]
to parameterise the parabola $\ds y^2=4kx$ we use $\ds x=kt^2,
\,\,\,\, y=2kt$


DIAGRAM


As $t$ increases this induces a direction on the curve.

The curve described in the opposite direction can be parameterised
by $\ds x=kt^2 \,\,\,\, y=-2kt$.


DIAGRAM


We regard these two as different curves (with the same set of
points).

It is important to distinguish the direction in many applications.
\end{itemize}


{\bf Polar equation of a conic}

We want to find the polar equation of a conic with the origin as
focus.


DIAGRAM

DIAGRAM


$\ds PS=ePM \,\,\,\, \sqrt{x^2+y^2}=e(x+k(e+1)) \hspace{0.5in}
(1)$

Converting to polars gives

$\ds r=er\cos\theta+ek(e+1)$

notice that from (1) $ ek(e+1)$ is the y-value when $x=0$


DIAGRAM


Write $l=ek(e+1)$.  The length $2l=PP'. \,\,\,\,  PP'$ is called
the latus rectum.  $l$ is the semi-latus rectum.

Thus we can write the conic as

$$ \frac{l}{r}=1-e\cos\theta$$

Note that rotations are easy in polar co-ordinates, so the
equation

$$ \frac{l}{r}=1-e\cos(\theta-\alpha)$$

is a conic having its axis at an angle $\alpha$ with the initial
line.  Notice that when $\alpha=\pi$ the equation becomes

$$ \frac{l}{r}=1+e\cos\theta$$

In the case of an ellipse or hyperbola this is equivalent to using
the other focus as an origin.

Notice that if $e&gt;1$ we can sometimes have $\ds \frac{l}{r}&lt;0.$
Although we normally insist on $r&gt;0$ in polars, in interpreting
polar equations it is often convenient to allow $r&lt;0$, meaning $r$
measured in the other direction through $O$.

e.g. $$\frac{1}{r}=1-2\cos\theta$$

when $\theta=0$ this gives $\ds \frac{1}{r}=-1, r=-1$.

We plot $\theta=0, r=-1$ as the point $(-1,0)$.

When $\ds \cos\theta=\frac{3}{4}, (\theta\approx 41^\circ)$ this
gives $\ds \frac{1}{r}=-\frac{1}{2}, r=-2$


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