intuition behind deriving the equation of a double-napped cone

Question

I know the equation of a double-napped cone is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2}$ but don't fully understand how this is derived. For a right circular cone centered at the origin, I know the equation is $x^2 + y^2 = z^2$. Is this because the double cone is a stack of different-sized circles, each with the equation $x^2 + y^2 = r^2$? I don't see how $r^2$ can be replaced with $z^2$ if the slope of the cone is not 1, and the slope of the cone could be anything.

If it's instead because the cone's surface is a set of hypotenuses of right triangles, the then the order of the variables wouldn't make sense, since a segment of length $z$ (in red above) would be a leg, not a hypotenuse. I'm also assuming that $a$, $b$, and $c$ deal with elliptical cones. Can anyone help me solve this conceptual puzzle?

Answer 1

Let $O$ be the origin and vertex of the circular cone, $P(x,y,z)$ be a point on it, $\theta$ be the semi-vertical angle and $(n_x,n_y,n_z)$ be a unit vector along the axis, then in the picture where half the disc perpendicular to the axis making a triangle with the angle $\theta$ at $O$ is the opposite, the hypotenuse is $\sqrt{x^2+y^2+z^2}$ and the adjacent is $n_x x+n_y y+n_z z$ (project $P$ to the axis) making $$\cos(\theta)=\frac{n_x x+ n_y y+n_z z}{\sqrt{x^2+y^2+z^2}}.$$ This gives an equation of the form $$(x^2+y^2)/a^2-z^2/c^2=0.$$

Now, the general cone $x^2/a^2+y^2/b^2-z^2/c^2=0$ lies between the two circular cones $(x^2+y^2)/a^2-z^2/c^2=0$ and $(x^2+y^2)/b^2-z^2/c^2=0,$ like an ellipse between the two circles on radii the semi-axes.

Answer 2

You are perfectly right with your first point: If the slope of the cone is not 1, we cannot simply replace $r^2$ by $z^2$. The equation $x^2+y^2=z^2$ is only the equation of a cone with slope 1.
Now how can we change that: If we replace $z$ by $\frac zc$, we scale the $z$ axis down by a factor of $c$. This has the effect of scaling the cone by a factor of $c$ in the $z$ direction, letting it be a cone of slope $c$. (If you can imagine this in 3D, try it out in 2D: start with the line $x-y=0$ and rescale $x$ and $y$.)
This works precisely the same in the other axis: The equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}$ is the equation of a cone with slope $c$, consisting of ellipses with eccentricity $\sqrt{1-\frac{a^2}{b^2}}$.

Answer 3

Let's first observe what $x^2+y^2 = z^2$ means. Fix some $z = r$. We get $x^2+y^2 = r^2$, so intersection of $x^2+y^2 = z^2$ and the plane $z = r$ is a circle of radius $r$ centered at origin.

We now want to go backwards, start with general ellipse in a plane with semi-axes $a$ and $b$. We have equation $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$.What we now want is to scale $a$ and $b$ with $z$ linearly to get a cone (bigger the $z$ is, bigger the semi-axes), so instead of semi-axes $a$ and $b$, let's have semi-axes $az$ and $bz$: $$\frac{x^2}{(az)^2}+\frac{y^2}{(bz)^2} = 1 \implies \frac{x^2}{a^2}+\frac{y^2}{b^2} = z^2$$ so we get equation of the cone such that its intersection with the plane $z = 1$ is ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$.

Finally, what is $c^2$ doing in your equation? Well, $\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{z^2}{c^2}$ is a scaled cone such that its intersection with the plane $z = c$ is ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$, i.e. ellipse with semi-axes $a$ and $b$. The $c$ is redundant, though, since $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{z^2}{c^2}\implies \frac{x^2}{(a/c)^2}+\frac{y^2}{(b/c)^2} = z^2.$$ Still, your equation might be nicer including $c$ as a parameter, since your cone is then determined by points $(0,0,0)$, $(a,0,c)$ and $(0,b,c)$ that lie on it, i.e. if you project your cone onto $xz$-plane, you get slopes $\pm c/a$ and if you project it onto $yz$-plane, you get slopes $\pm c/b$. Of course, the point is that if you have rational slopes, you can easily choose positive integers $a,b,c$ to write the equation of the corresponding cone.

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Knowing the correct equation, you can also check directly that it works: $$ \frac{x^2}{a^2}\!\!+\!\frac{y^2}{b^2} \!=\! \frac{z^2}{c^2} \!\!\implies\!\! \left(\frac{cx}{az}\right)^2 \!\!+\! \left(\frac{cy}{bz}\right)^2 \!\!\!=\! 1 \!\!\implies\!\! \frac{cx}{az} \!=\! \cos t, \frac{cy}{bz} \!=\! \sin t \!\!\implies\!\! x \!=\! \frac{az}{c}\cos t, y \!=\! \frac{bz}{c}\sin t$$ so we can see that for any fixed $z$, $(\frac{az}{c}\cos t,\frac{bz}{c}\sin t),\ t\in[0,2\pi]$ is an ellipse.