Introduction
Quadric surfaces are of the form \(Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\), where the sum of the powers of any term is less than or equal to 2. However, we don't usually see them in this form. Translations combine the \(x^2\) and \(x\) terms; we only see linear terms if they appear without the cooresponding squared term. Additionally, the \(xy\), \(xz\), and \(yz\) terms are eliminated with rotations. We can use translations and rotations to turn the given equation into a more intuitive form because shapes are the same regardless of where they're located or how they're oriented.
There are seven standard quadric surfaces to consider.
1. Ellipsoid \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\)
2. Cone \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0\)
3. Hyperboloid of One Sheet \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\)
4. Hyperboloid of Two Sheets \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = -1\)
5. Elliptic Paraboloid \(z=\frac{x^2}{a^2} + \frac{y^2}{b^2}\)
6. Hyperbolic Paraboloid \(z=\frac{x^2}{a^2} - \frac{y^2}{b^2}\)
7. Cylinder \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = r\)
When intersected by a plane, we see conic sections! These can be hyperbolas, parabolas, ellipses, circles, or a combination of these depending on how many times the surface is sliced. Clearly, some of these appear more intriguing than others. A slice that reveals only a line, point, or pair of lines isn't all that interesting. (Source: https://www.youtube.com/watch?v=VAz-X6uaSs4)
The surface and plane I chose
Consider a cone centered at the origin given by \(\frac{z^2}{144}=\frac{x^2}{25}+\frac{y^2}{4}\).
I chose this function because cones show every conic section and are most interesting to slice with a plane! I picked values for a, b, and c such that the opening of the cone is an ellipse and not a circle, just to spice things up.
Now, let's slice it. I want to show a hyperbola because these are the coolest conic sections in my opinion. To do this, I want a vertical plane like the one pictured. (Source: https://math.libretexts.org)
The equation of a plane is given by \(Ax+By+Cz+D=0\). To find the equation of a plane, all you need is a point and a normal vector. For example, choose the plane perpendicular to (1, -1, 0). Points perpendicular to a vector \(\vec{v}=\left\lceil\begin{matrix} v_{1}\\v_{2}\\v_{3} \end{matrix}\right\rceil\) are the points \(\vec{x}=\left\lceil\begin{matrix} x_{1}\\x_{2}\\x_{3} \end{matrix}\right\rceil\) such that \(\vec{v}\cdot\vec{x}=0\). So, we do the following:
\[1\cdot x+(-1)\cdot y+0\cdot z=0\]
\[x-y=0\]
\[x=y.\]
Here's the cone sliced with the plane \(x=y\):
I like how it's cut so that the ellipse-like opening is at a slant.
The intersection
Now for the math! We are intersecting \(\frac{z^2}{144}=\frac{x^2}{25}+\frac{y^2}{4}\) with the plane \(x=y\). To do this, we can use a really simple substitution. Set \(x\) equal to \(y\) to obtain \[\frac{z^2}{144}=\frac{x^2}{25}+\frac{x^2}{4}\] \[z^2=144[x^2(\frac{1}{25}+\frac{1}{4})]\] \[z^2=144(0.29x^2)\] \[z^2=41.76x^2\] \[z=\pm 6.46x\] On the 3D print, we can see the hyperbola by pulling the two halves of the model apart. The model is just under 4 inches tall, so \(z\) is from around -2 to 2. It is about one inch at its widest, where \(y\) ranges from around -0.5 to 0.5.
Hopefully now you can see why cones are the coolest surface and hyperbolas are the most interesting cross-section. Our cone may look more like a squished hourglass than any cone we're used to seeing, but that's okay! If we only wanted to see one half of this hourglass shape, we would have chosen
\[z^2=c^2(\frac{x^2}{a^2}+\frac{y^2}{b^2})=\frac{c^2}{a^2}x^2+\frac{c^2}{b^2}y^2=A^2x^2+B^2y^2\]
\[z=\pm\sqrt{A^2x^2+B^2y^2}.\]
Additionally, our cone opens up along the z-axis. If we wanted to change that, we would only need to change our equation slightly. For example, a cone opening along the x-axis would be given by \(\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x^2}{a^2}.\) (Source: https://tutorial.math.lamar.edu/Classes/CalcIII/QuadricSurfaces.aspx)
Word count: 522
Quadric surfaces are of the form \(Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\), where the sum of the powers of any term is less than or equal to 2. However, we don't usually see them in this form. Translations combine the \(x^2\) and \(x\) terms; we only see linear terms if they appear without the cooresponding squared term. Additionally, the \(xy\), \(xz\), and \(yz\) terms are eliminated with rotations. We can use translations and rotations to turn the given equation into a more intuitive form because shapes are the same regardless of where they're located or how they're oriented.
There are seven standard quadric surfaces to consider.
1. Ellipsoid \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\)
2. Cone \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0\)
3. Hyperboloid of One Sheet \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1\)
4. Hyperboloid of Two Sheets \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = -1\)
5. Elliptic Paraboloid \(z=\frac{x^2}{a^2} + \frac{y^2}{b^2}\)
6. Hyperbolic Paraboloid \(z=\frac{x^2}{a^2} - \frac{y^2}{b^2}\)
7. Cylinder \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = r\)
When intersected by a plane, we see conic sections! These can be hyperbolas, parabolas, ellipses, circles, or a combination of these depending on how many times the surface is sliced. Clearly, some of these appear more intriguing than others. A slice that reveals only a line, point, or pair of lines isn't all that interesting. (Source: https://www.youtube.com/watch?v=VAz-X6uaSs4)
Consider a cone centered at the origin given by \(\frac{z^2}{144}=\frac{x^2}{25}+\frac{y^2}{4}\).
Now, let's slice it. I want to show a hyperbola because these are the coolest conic sections in my opinion. To do this, I want a vertical plane like the one pictured. (Source: https://math.libretexts.org)
The intersection
Now for the math! We are intersecting \(\frac{z^2}{144}=\frac{x^2}{25}+\frac{y^2}{4}\) with the plane \(x=y\). To do this, we can use a really simple substitution. Set \(x\) equal to \(y\) to obtain \[\frac{z^2}{144}=\frac{x^2}{25}+\frac{x^2}{4}\] \[z^2=144[x^2(\frac{1}{25}+\frac{1}{4})]\] \[z^2=144(0.29x^2)\] \[z^2=41.76x^2\] \[z=\pm 6.46x\] On the 3D print, we can see the hyperbola by pulling the two halves of the model apart. The model is just under 4 inches tall, so \(z\) is from around -2 to 2. It is about one inch at its widest, where \(y\) ranges from around -0.5 to 0.5.
Word count: 522
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