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

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.

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.