\section{Quadric Surfaces}
\textbf{Quadric surfaces} are surfaces which are described by $2$nd order equations in the variables $x,y,z$. An equation is of the \textbf{second order} if it is polynomial in each of its variables and has total degree two. (For example, $x^2+y+xz-3z^2=6$ is an equation of the second order, but $x^2y^2-z=0$ is not.)

Quadric surfaces provide interesting geometric examples. A major reason for us to be interested in quadric surfaces is that we will use them to approximate surfaces with more complicated equations.

Each of the following subsections introduces one type of quadric surfaces in $\mathbb{R}^3$.  They are: spheres/ellipsoids, cylinders, paraboloids, hyperboloids and cones. We give a short description of each, with accompanying computer-rendered images. We are not attempting a complete classification of quadric surfaces (although that can be done---classifying surfaces given by various types of algebraic equations is a major area of study in algebraic geometry). Instead, our focus will be on recognizing---and working with---certain basic types of surfaces that will be useful in the sequel.


\subsection{Spheres and Ellipsoids} These surfaces are defined by the equations:
\begin{align*}
(\textrm{Sphere}) & \quad & x^2 + y^2 +z^2 = r^2 & \\
(\textrm{Ellipsoid}) & \quad & \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
\end{align*}
where $a,b,c,r$ are all real, non-zero numbers. Note that the equations above do not change if we replace $r$ by $-r$, $a$ by $-a$, and so on; for convenience, we will always assume that $a,b,c,r>0$.  A sphere is actually a special kind of ellipsoid, with $a = b = c = r$ for some $r > 0$. For an ellipsoid centred at some point $(x_0, y_0,z_0)$ instead of the origin $(0,0,0)$, we have the equation
$$
\frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} + \frac{(z - z_0)^2}{c^2} = 1.
$$
The other surfaces described below can be translated and rescaled in a similar way.


\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig1}
\centering
\includegraphics[width= .7\textwidth]{15fig1.png}
\caption{An ellipsoid centred at the origin.}
\end{figure}
\end{minipage}



\subsection{Cylinders} The ``usual" circular cylinder about the $z$-axis has the equation
\begin{align*}
(\textrm{Circular Cylinder}) & \quad & x^2 + y^2 = r^2
\end{align*}
Notice that the $z$ variable does not make an appearance in the equation defining the circular cylinder! This means that the surface is \textit{invariant under translations} parallel to the $z$-axis: if we move the surface by a vector parallel to the $z$-axis, we get the same surface. Additionally, the circular cylinder is also symmetric about the $z$-axis. 


\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig2}
\centering
\includegraphics[width= .7\textwidth]{15fig2.png}
\caption{A circular cylinder about the $z$-axis with $r = 3$. }
\end{figure}
\end{minipage}

\medskip

The dark curve in the previous figure is given by the system of equations
$$
\begin{cases}
x^2 + y^2 = 3 \\
z = 1
\end{cases}
$$
These are the equations for a circle of radius $3$ and the equation for the plane $z = 1$.  In words,  the intersection of the circular cylinder with a plane parallel to the $xy$-axis is the circle $x^2 + y^2 = r^2$. Hence the name---\textit{circular} cylinder.


We can also consider elliptic and parabolic cylinders.  These are given by the equations
\begin{align*}
(\textrm{Elliptic Cylinder}) & \quad & \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\
(\textrm{Parabolic Cylinder}) & \quad & \frac{x^2}{a^2} - y = 0
\end{align*}
Notice that, just like the circular cylinder, both of these equations omit the $z$ variable.  This means that these surfaces are also invariant under translations parallel to the $z$-axis.  Moreover, the intersection of these surfaces with any plane parallel to the $xy$-axis is an ellipse (resp. parabola).


\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig3}
\centering
\includegraphics[width= .7\textwidth]{15fig3.png}
\caption{A parabolic cylinder about the $z$-axis. }
\end{figure}
\end{minipage}


\bigskip

\subsection{Paraboloids} In these surfaces, one variable occurs only in the first power, so that the equation is of order one in that variable.  Here are some examples of paraboloids where the $z$-variable occurs only in the first power:
\begin{align*}
(\textrm{Elliptic Paraboloid}) & \quad & z = \frac{x^2}{a^2} + \frac{y^2}{b^2} \\
(\textrm{Hyperbolic Paraboloid}) & \quad & z = \frac{y^2}{b^2} - \frac{x^2}{a^2}
\end{align*}



\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig4}
\centering\includegraphics[width= .7\textwidth]{15fig4.png}
\caption{An elliptic paraboloid oriented about the $z$ axis. }
\end{figure}
\end{minipage}


\bigskip

Similar to the case of spheres and ellipsoids, we give the case $a = b$ a special name: this equation describes a \textbf{circular} paraboloid.  Recall that a circle is a special case of an ellipse.

\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig5}
\centering\includegraphics[width= .7\textwidth]{15fig5.png}
\caption{A hyperbolic paraboloid. }
\end{figure}
\end{minipage}

\bigskip

Figure 5 shows the hyperbolic paraboloid intersecting the $xy$ plane in a pair of straight lines. This can be confirmed using the equation $z = \frac{y^2}{b^2} - \frac{x^2}{a^2}$ above: setting $z=0$, we get $\frac{y^2}{b^2} - \frac{x^2}{a^2}=0$, meaning that either $\frac{y}{b}=\frac{x}{a}$ or $\frac{y}{b}=-\frac{x}{a}$. These are the two lines you see in the $xy$-plane.

What's less clear from the picture is that a hyperbolic paraboloid actually contains many more straight lines! Let us show this algebraically.  First, rewrite the equation $z = \frac{y^2}{b^2} - \frac{x^2}{a^2}$ as
$$
z = \left(\frac{y}{b} + \frac{x}{a} \right)\left(\frac{y}{b} - \frac{x}{a}  \right).
$$
Now, fix a real number $c \neq 0$, and consider the line $L$ at the intersection of the two planes
$$
\begin{cases}
\frac{y}{b} + \frac{x}{a} = \frac{z}{c} \\
\frac{y}{b} - \frac{x}{a} = c
\end{cases}
$$
(Exercise: check that these two planes are not parallel!) 
Any point $P=(x,y,z)$ on the line $L$ has to satisfy both of these equations. Multiplying out, we get
$$
\left(\frac{y}{b} + \frac{x}{a} \right)\left(\frac{y}{b} - \frac{x}{a}  \right) = \frac{z}{c} \cdot c = z
$$
so that any point $P=(x,y,z)$ on the line $L$ must also satisfy the equation defining the hyperbolic paraboloid. Therefore, these lines are necessarily contained in the hyperbolic paraboloid, regardless of our choice of $c$.

Similarly, we can get \textit{another} family of lines contained in the hyperbolic paraboloid by looking instead at the equations
$$
\begin{cases}
\frac{y}{b} + \frac{x}{a} = c \\
\frac{y}{b} - \frac{x}{a} = \frac{z}{c}
\end{cases}
$$
This is an example of a \textit{doubly ruled surface}: for every point $P$ on the hyperbolic paraboloid, there are two distinct lines passing through $P$ that are contained in the surface. (\underline{Exercise: prove this!} Most of the work was already done. What is left to do is check that, for every point $P$ on the paraboloid, two of the lines listed above pass through $P$.)



\bigskip


\subsection{Hyperboloids} There are two types of hyperboloids, given by the equations:
\begin{align*}
(\textrm{Hyperboloid of One Sheet}) & \quad & \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \\[1ex]
(\textrm{Hyperboloid of Two Sheets}) & \quad & \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = -1
\end{align*}

The term \textit{sheet} here refers to whether the hyperboloid splits into one piece or two pieces in space.  You can see this readily in the following figures.


\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig6}
\centering
\includegraphics[width= .7\textwidth]{15fig6.png}
\caption{A hyperboloid of one sheet. }
\end{figure}
\end{minipage}

\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig7}
\centering
\includegraphics[width= .7\textwidth]{15fig7.png}
\caption{A hyperboloid of two sheets.}
\end{figure}
\end{minipage}


\bigskip

What do these surfaces look like ``at infinity''? Suppose $a = b = 1$,  and rewrite the above equations as
$$
c^2 (x^2 + y^2) = z^2 \pm c^2.
$$
Notice that, in this equation, if $|z|$ is much, much larger than $|\pm c| = |c|$,  then the surface should look a lot like whatever surface is described by the equation
$$
c^2 (x^2 + y^2) = z^2.
$$
This is because the $\pm c^2$ does not wiggle the value of $z$ too much. This tells us that, away from the origin (where $z$ is large) the hyperboloids look more and more like our next surface.


\bigskip

\subsection{Cones} A (circular) cone is described by the equation $z^2 = c^2 (x^2 + y^2)$.  

\begin{minipage}{.8\textwidth}
\begin{figure}[H]\label{15fig8}
\centering
\includegraphics[width= .7\textwidth]{15fig8.png}
\caption{A circular cone.}
\end{figure}
\end{minipage}

\bigskip

Notice that the equation with the $z$ variable in second-order covers both
$$
z = + c \sqrt{x^2 + y^2} \quad \textrm{and} \quad z = - c \sqrt{x^2 + y^2},
$$
simultaneously.  If we only want to include one part of the cone, we would choose one of the above equations. With $c>0$, the $+$ corresponds to the upper cone and the $-$ corresponds to the lower cone. There are also notions of an elliptic and parabolic cone, which we will not cover here.  Try to use a graphing software like Geogebra to discover what these equations would look like!


\subsection{Identifying Surfaces and Their Corresponding Equations}
Often, we are presented with an image of some surface and would like to match it to a given equation.  It can then be helpful to ask the following questions.

\begin{itemize}
\item Is the surface \textbf{bounded} or \textbf{unbounded}? In other words: are the variables $x,y,z$ allowed to take on infinitely large values (meaning unbounded)? As an example: every sphere or ellipsoid is bounded, but the paraboloid surfaces we discussed are unbounded. Can you think of why this must be true?

\smallskip

\item Are there any \textbf{intercepts} with the $x$- or $y$- or $z$-axes? For example, the cone intersects all three axes at the origin, because when $x = y = 0$, we necessarily have $z = 0$ (and similarly for any two variables simultaneously equalling zero).  However, the hyperboloid of one sheet never intersects the $z$-axis.  Try to figure out why this is true! Hint: remember that the $z$ axis is given by the equations $x=y=0$.


\smallskip

\item Does the surface \textbf{intersect} the planes $x = 0$ or $y = 0$ or $z = 0$? What do these intersections look like?
You can also check and see how the surface intersects the planes $x = c$ or $y = c$ or $z = c$ for various values of $c$. Be creative and try to pay attention to any special values appearing on the variables $x,y,z$ for hints. These intersections with planes are called \textbf{traces} of the surface.

\smallskip

\item Does the surface have any symmetries? For example, is it a surface of revolution? Is this reflected in the equations you are given? (Exercise: which of the surfaces in this section are surfaces of revolution?)


\end{itemize}

%\vfill
%\pagebreak

\section*{Worked Problems}

{\it 1. Pair the accompanying figures to the following two equations:}
$$
z = -xy \quad \quad \textit{and} \quad \quad z^2= x^2 + y^2 - 1.
$$
\medskip

\begin{figure}[!htb]
    \centering
    \begin{minipage}{.5\textwidth}
        \centering
        \includegraphics[width=0.45\linewidth, height=0.15\textheight]{15fig9.png}
        \caption{}
    \end{minipage}%
    \begin{minipage}{0.5\textwidth}
        \centering
        \includegraphics[width=0.45\linewidth, height=0.15\textheight]{15fig10.png}
        \caption{}
    \end{minipage}
\end{figure}

%\pagebreak

\bigskip
The correct answer is that Figure 1.21 matches $z = -xy$ and Figure 1.22 matches $z^2= x^2 + y^2 - 1 $.   There are several ways we could show this:

\medskip

\begin{itemize}
\item The origin $(0,0,0)$ belongs to the surface in Figure 1.21 but not in Figure 1.22.  Notice that $x=y=z=0$ is a solution to the equation $z = -xy$ but not the equation $x^2 + y^2 - 1 = z^2$.

\item Figure 1.21 has an unrestricted range of values for $x$ and $y$.  However,  Figure 1.22 shows that $x$ and $y$ are bounded away from the origin.  This matches the fact that $x^2 + y^2 = z^2 + 1 \geq 1$ (since $z^2 \geq 0$).

\smallskip

\item The intersection of Figure 1.22 with any plane $z = c$ is a circle.  This matches the equation $x^2 + y^2 = c^2 +1$--an equation for a circle of radius $\sqrt{c^2 + 1}$.  
(This also gives Figure 1.22 a rotational symmetry about the $z$-axis, since circles are symmetric under rotation.)

\smallskip

\item For another symmetry: make the change of variables $z \mapsto -z$.  Then,  $x^2 + y^2 -1 = (-z)^2 = z^2$ but $-z = -(-xy) = xy$.  In other words,  only the equation $x^2 + y^2 -1 = z^2$ is symmetric under the reflection $z \mapsto -z$.  This matches Figure 1.22 (since if we flip the surface upside-down, we get the same surface).

\smallskip

\item Take the plane passing through the origin and parallel to the sheet of paper you are reading.  In the coordinate system indicated in Figures 1.21 and 1.22, this is the plane $x=-y$. 

\smallskip
\noindent
From the geometric representation, the intersection of this plane with the surface in Figure 1.21 is a parabola, whereas its intersection with the surface in Figure 1.22 is a hyperbola. Now let's see what happens if we try to compute the intersections with the same plane using the given equations. Making the substitution $x=-y$ in both equation gives:
$$
z = -(-y)y = y^2 \quad \quad \textrm{and} \quad \quad 2x^2 - 1 = z^2.
$$
These are the equations of a parabola and hyperbola (respectively). This, again, matches Figure 1.21 to $z = -xy$ and Figure 1.22 to $x^2 + y^2 - 1 = z^2$


\end{itemize}




