Is the right intersection of an oblique circular cone an ellipse?

Question

The bottom of a glass cup (assuming it is a true circle) looks like & is drawn as an ellipse. But is the shape we are seeing really an ellipse?

After some tinkering with pen and paper, I see that this question comes down to: Is the right intersection of an oblique circular cone an ellipse?

*edit
@Aretino: I am certain my question concerns an oblique circular cone. I'm sorry my example with the glass cup was a little too specific and not explained fully. Here's another fuller explanation.

Standing afar a circle drawn on the ground, you see an ellipse-like shape. Here the rays of light coming from the circle on the ground into your eye form an oblique circular ellipse. Supposing that the eye works like a camera, the image we see is a right intersection of the cone of incoming light rays.

Answer 1

Suppose we have a circle of diameter $MN$ lying in a plane $\alpha$, and a point $V$ outside $\alpha$, which is the vertex of an oblique cone having circle $MN$ as its guiding line. I'll show below, following the derivation made by Apollonius of Perga in his treatise on conic sections, that the intersection between this cone and a plane can be described, using cartesian coordinates, by the usual equation of an ellipse.

Let's choose diameter $MN$ so that plane $VMN$ is perpendicular to $\alpha$. The cone is cut by another plane $\beta$, also perpendicular to $VMN$, intersecting the cone along a curve $APB$ (red in diagram below), where $A$ and $B$ are in particular the intersections of $\beta$ with lines $VM$ and $VN$. The more general case can be treated in a similar way, but the equation is then obtained with respect to a pair of oblique coordinate axes.

Let $P$ be any point on the curve and draw a plane through $P$ parallel to $\alpha$: it is easy to show that its intersection with the cone is a circle. Let $QR$ be the diameter of the circle parallel to $MN$. The perpendicular $PH$ from $P$ to $AB$ is also perpendicular to $MN$ and we have, by similitude in right triangle $QPR$: $$ PH^2=QH\cdot RH. $$ Draw now from $A$ and $B$ lines $AE$ and $BD$, parallel to $QR$. Triangles $QHA$ and $RHB$ are similar to $DBA$ and $EAB$, hence $QH=(BD/AB)AH$ and $RH=(AE/AB)BH$. Substituting that into the previous equation we get: $$ PH^2={BD\cdot AE\over AB^2}AH\cdot BH. $$ Finally, if $C$ is the midpoint of $AB$ we can write: $AH=AC+CH$, $BH=AC-CH$. With this substitution our equation becomes: $$ PH^2={BD\cdot AE\over 4AC^2}(AC^2-CH^2), $$ which can also be written as: $$ {CH^2\over AC^2}+{4PH^2\over BD\cdot AE}=1. $$ If we set up, in plane $\beta$, a coordinate system centred at $C$ and with $x$ axis along $AB$, then $CH$ and $PH$ are the coordinates of point $P$ and the above equation can be written as $$ {x^2\over a^2}+{y^2\over b^2}=1, \quad\text{where:}\quad a=AC,\quad b=\sqrt{BD\cdot AE}/2, $$ which is the usual equation of an ellipse with semi-axes $a$ and $b$ in cartesian coordinates.

Answer 2

The intersection of any quadric surface in $\mathbb R^3$ is a (perhaps degenerate) conic. This is fairly easy to show analytically using homogeneous coordinates.

Let $Q$ be a symmetric $4\times 4$ real matrix. This matrix defines some quadric surface via the equation $\mathbf X^TQ\mathbf X=0$ (where $\mathbf X=(x,y,z,1)^T)$. Further, let the $4\times 3$ matrix $M$ define a coordinate system on the intersecting plane via the mapping $\mathbf X=M\mathbf x$. Substituting into the quadric equation, we get $$(M\mathbf x)^TQ(M\mathbf x) = \mathbf x^T(M^TQM)\mathbf x = 0.$$ The matrix $M^TQM$ is a symmetric $3\times3$ real matrix, so this is the equation of a conic in the plane.

So, if the curve of intersection of your plane and cone looks like an ellipse, it is an ellipse.

Another simple way to see this is that your situation is related to the canonical conic section construction via some nonsingular affine transformation, and the image of an ellipse under such a transformation is another ellipse.