I did not even use calculus at all; I don't even know calculus, nor do I know if calculus is usable to find the formula. I am in a geometry class right now, for I am a high school freshman.
But still, I might have two different solutions for different kinds of oblique cones.
I originally tried to share the actual document, since the formulas contain characters that can't be used for this inspection, but according to James, the document was not accessible. Unfortunately, I do not know how exactly to share the document to everyone, so I just took a screenshot of the document and all it has to offer.
[Link of Screenshot][1] [1]: https://i.stack.imgur.com/91gIu.png
I also tried to explain everything to answer James's other questions
An oblique cylinder is easier since it has an entirely congruent slant height, which is congruent to the axis. So an accurate lateral area formula would be 2πra, where r is the radius and a is the axis. And what I mean by "axis" is the distance between the center of the base and the vertex.
However, an oblique cone only has one pair of congruent slant heights that connect to points that are the base's diameter apart. That slant height can actually be interpreted as the mean of the maximum and minimum slant heights, in which is the average slant height of the entire cone. This is so because the increase/decrease in the slant height from the two paired slant heights mentioned previously (which this is why they are important) are exactly the same since the base is a circle and not a parabola.
To support this, those points create a diameter that is perpendicular to the diameter connecting the points of the maximum and minimum slant heights.
The lateral area found only using the Pythagorean Theorem twice in a row, only being given the radius, the altitude, and the axis, besides the fact that an angle measurement of the axis can replace either the axis or the altitude, because trigonometry, and after using the trigonometry is when the formulas are usable. It's also what is multiplied by πr to find the lateral area, just like a right cone, for the surface area, that is simply one of the formulas + the base: πr^2.
Let me know if there is anything wrong with having access to the screenshot, either if the image is not public, or the stupid fact that I do not have enough reputation.
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.
Best Answer
In this answer, I derive a formula for the eccentricity of an ellipse in terms of angles made by the surface of the cone, and the cutting plane, with a "horizontal" plane (perpendicular to the axis). Complementing (for the sake of easier description here), we can restate the formula as $$e = \frac{\cos\theta}{\cos\phi} \tag1$$ where $\phi$ is the angle made by (a generator of) the cone and its axis, and $\theta$ is the (non-obtuse) angle made by the cutting plane (or, more-specifically, the conic's major axis) and the cone axis. It easy to show that, for the case of a cylinder, the corresponding relation matches the result of setting $\phi=0$ (and writing $\theta'$ to distinguish from $\theta$): $$e = \cos\theta' \tag2$$ So, if a cone and cylinder have a planar (thus, elliptical) intersection, then the ellipse's major axis makes angles $\theta$ and $\theta'$ with their respective axes, such that $$\cos\theta'=\frac{\cos\theta}{\cos\phi} \tag3$$ Thus, given any plane cutting a cone in an ellipse, we can find an appropriate cylinder that has the ellipse as its intersection. (Technically, the ellipse may only be part of the overall intersection.) Knowing $\phi$ and $\theta$, we use $(3)$ to find $\theta'$. (Note: Given that the plane cuts the cone in an ellipse, we know $1>e=\cos\theta'$ so that $\theta'$ is defined.) This angle determines the cylinder's axis through the ellipse's center; in fact, for $\theta'\neq \pi/2$, we have two possibilities for that axis. The desired cylinder's radius is necessarily the ellipse's minor radius. So, generally, there's are two families of configurations (essentially parameterized by $\theta$ and a scale factor, and the choice of cylinder axis) of cones and cylinders having (partially) planar intersection, with no requirement that the cone and cylinder have parallel axes.
That said, in the particular case where the cone and cylinder do have parallel axes, then $\theta=\theta'$. Assuming $\phi\neq 0$ (the cone is non-degenerate), then the only way to satisfy $(3)$ is with $\cos\theta=\cos\theta'=0=e$: the intersection is a circle. Clearly, the cylinder's axis necessarily coincides with the cone's. An "offset" configuration ($a\neq 0$ in the question) is invalid, re-confirming conclusions made in other answers.