conic sectionsgeometry
As shown in the figure above, I want to determine the equation of the ellipse formed by intersection of a tilted right cone and a plane. We know $\alpha,$ (angle formed with the vertical), $R$ as well as $\Omega$ (cone's half-angle). The author here Previously given solution proposed a solution which confuses me on the $x=(p_1-p_2)/2,$ $y=R\tan(\Omega)$ part…how do we know this point lies on the ellipse? The author did not justify this step clearly.
Converting comments (about the original version of the question) to answer, as requested.
The cutting plane should be tangent to the Dandelin spheres. For the hyperbola case, looking from the side (as in your image), the "cutting line" should be externally tangent to the two "Dandelin circles".
It may be easier, in all cases, to start with the cone and cutting plane. Again, looking from the side, the "cone lines" and "cutting line" will determine, in the cone's interior, certain regions bounded by three sides (segments and/or rays). For an ellipse, there will be a finite triangle and an unbounded three-sided region; for a hyperbola, there will be two unbounded three-sided regions; for a parabola, there will be one unbounded three-sided region. Each "Dandelin circle" is tangent to the three sides of such a region.
Here's "unified" picture of a Dandelin configuration for both ellipses and hyperbolas (and parabolas as a limiting case). It consists simply of two circles, $\bigcirc K$ and $\bigcirc K'$, and their common tangents: the internal tangents meet at $A$, making an angle $\alpha$ with the line of centers; and the external tangents meet at $B$, making an angle $\beta$ with the line of centers. Necessarily, $\alpha > \beta$. (Exploration of the case where $\alpha=\beta$, which defines a parabola, is left to the reader.)
As indicated, $V=W$, $V'$, and $W'$ are points where an internal tangent meets an external. Internal tangent $\overleftrightarrow{VV'}$ touches the circles at $F$ and $F'$; external tangent $\overleftrightarrow{WW'}$ touches the circles at $G$ and $G'$. Note: because $|VF'|=|WG'|$ and $|VF|=|WG|=|V'F'|=|W'G'|$ (verification left to the reader), we have $|VV'|=|GG'|$ and $|WW'|=|FF'|$.
This allows us to restate the calculations made by OP:
It's worth reiterating that $(1)$ and $(2)$ say exactly that eccentricity of a conic is simply a ratio of internal and external tangent segments for these "Dandelin circles". I don't believe this is a new or particularly-profound observation, but I personally have never thought of things in quite this way. Nifty!
Also, as mentioned by OP, we have these relations involving the Dandelin radii, $r := |KF|=|KG|$ and $r':=|K'F'|=|K'G'|$: $$r'+r = |KK'|\sin\alpha \qquad r'-r=|KK'|\sin\beta$$
These imply $$\begin{align} r &= \frac12|KK'|(\sin\alpha-\sin\beta) = |KK'| \cos\frac12(\alpha+\beta)\sin\frac12(\alpha-\beta) \\[4pt] r' &= \frac12|KK'|(\sin\alpha+\sin\beta) = |KK'| \sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta) \\ \end{align}$$ so that $$\frac{r'}{r} = \frac{\tan\frac12(\alpha+\beta)}{\tan\frac12(\alpha-\beta)} \tag{3}$$
Here is an approach that avoids dividing into cases as well as any trigonometry. Suppose that the directions $d^1 = (d^1_1,d^1_2)$ and $d^2 = (d^2_1,d^2_2)$ are given in clockwise order. Rotate these vectors by $90^\circ$ clockwise and counterclockwise respectively to product $$ l = (d_2^1,-d^1_1), \quad r = (-d^2_2, d^2_1). $$ A point $x = (x_1,x_2)$ will lie within the cone if and only if it satisfies $x \cdot l \geq 0$ and $x \cdot r \geq 0$, where $v \cdot w$ denotes the dot-product of vectors $v$ and $w$. More specifically, we have $x \cdot l \geq 0$ iff $x$ lies to the "right" of the "left-side" boundary, and $x \cdot r \geq 0$ iff $x$ lies to the "left" of the "right-side" boundary.
We are given two endpoints $p^1 = (p^1_1,p^1_2)$ and $p^2 = (p^2_1,p^2_2)$. The line connecting these points is the set of all points $$ p(t) = (1-t)p^1 + tp^2 $$ with $t \in \Bbb R$. Note that $p(t)$ is on the line segment connecting the two points when $0 \leq t \leq 1$. Moreover, $p(0) = p^1$ and $p(1) = p^2$.
We now find the "times" $t$ at which this line crosses either of the boundaries. That is, we solve $$ l \cdot p(t_l) = 0 \implies (1-t_l)(l \cdot p^1) + t_l(l \cdot p^2) = 0 \implies t_l = \frac{l \cdot p^1}{(l \cdot p^1) - (l \cdot p^2)},\\ l \cdot p(t_r) = 0 \implies (1-t_r)(l \cdot p^1) + t_r(l \cdot p^2) = 0 \implies t_r = \frac{r \cdot p^1}{(r \cdot p^1) - (r \cdot p^2)}. $$ If either of these numbers satisfy $0 \leq t \leq 1$, plug into $p(t)$ to produce the associated point.
The only case not accounted for here is division by zero, which occurs when the line segment is parallel to one of the boundaries.
Best Answer
Here's a 3-D picture, to explain more clearly that answer.
I chose as $x$-axis the intersection between the given plane, and a plane through the axis of the cone, perpendicular to it. I placed the origin at the midpoint of segment $AB$ (which is the major axis of the ellipse). $y$-axis is on the given plane, perpendicular $x$-axis. If you draw, from the center $M$ of the cone base, a line parallel to $y$-axis, then it intersects the cone (and the ellipse) at points $I$ and $H$, with coordinates $|x|=CM=(𝑝_1−𝑝_2)/2$, $|y|=MH=𝑅\tan Ω=\,$radius of the base of the cone.
EDIT.
To choose the intersecting plane such that the center of the ellipse is at a given point $C$ inside the cone, let's consider line $VC$, where $V$ is the vertex of the cone (see figure below).
If $C$ lies on the axis of the cone, then just choose a plane through $C$ perpendicular to the axis, to obtain as intersection a circle of centre $C$.
If $C$ doesn't lie on the axis, then consider the plane $\sigma$ through $C$ and the axis, intersecting the cone at two generatrices $VA$ and $VB$, and set $\angle CVB=\beta$, $\angle CVA=\gamma$, with $\beta>\gamma$. If a plane through $C$, perpendicular to $\sigma$, intersects $VA$, $VB$ at $A$ an $B$ respectively, then $AB$ is the major axis of the intersection ellipse, and $C$ is its center if $AC=BC$.
Let $\theta=\angle VCB$. From the sine law applied to triangles $VCB$ and $VCA$ one finds
$$ {BC\over VC}={AC\over VC}={\sin\beta\over\sin(\theta+\beta)}= {\sin\gamma\over\sin(\theta-\gamma)}, \quad\text{whence:}\quad \tan\theta={2\sin\beta\sin\gamma\over\sin(\beta-\gamma)}. $$ Finally, we can express angle $\alpha$ between the plane of the ellipse and a plane perpendicular to the axis as a function of $\theta$: $$ \alpha={\pi+\gamma-\beta\over2}-\theta. $$