Locating focus and semi latus-rectum of conic by R&C construction.

conic sectionsgeometric-constructionsolid-geometry

A cone semi-vertex angle $\gamma$ is cut by a plane inclined at angle $\alpha$ to symmetry axis. Length of perpendicular on plane from the cone vertex is $q$.

It is known that eccentricity

$$ \epsilon =\frac{\cos\alpha}{\cos\gamma} $$

can be calculated. But can the semi latus-rectum $p$ or its length $\dfrac {q}{p}=f(\alpha,\gamma)$ be represented/found by geometric construction?

Thanks in advance for suggestion of a simple Ruler/Compass construction.

Finding foci for hyperbola intersection drawing direct tangents of Dandelin spheres.. is this how it is done?

Best Answer

If $V$ and $W$ are the vertices of the conic section, and $VV'$, $WW'$ are perpendicular to the axis of the cone, then the latus rectum is given by (see here for details): $$ 2p={VV'\cdot WW'\over VW}. $$ This is valid for an ellipse or hyperbola: the equation for a parabola can be recovered by letting $OW\to\infty$, where $O$ is the cone vertex.

EDIT.

See below for a ruler-and-compass construction of foci $F$ and $G$, based on the formula for the distance between a focus and the center of the ellipse $c={1\over2}\sqrt{VW^2−VV'\cdot WW'}$.

Related Solutions

[Math] Evaluate the eccentricity of the elliptical section of a right circular cone

Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $\theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $\beta$ so that $\beta = \alpha/2$ (this will help simplify the calculations later).

Clearly in $\Delta AOC$ we have $$\frac{OA}{\sin \angle OCA} = \frac{AC}{\sin \angle AOC}$$ so that $$b = AC = \dfrac{\sin \theta}{\sin(\theta + \beta)}\tag{1}$$ Similarly $$c = AB = \frac{\sin \theta}{\sin(\theta - \beta)}\tag{2}$$ and $$a = BC = \frac{b\sin 2\beta}{\sin(\theta - \beta)} = \frac{\sin\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}\tag{3}$$ Let the incircle of $\Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $\Delta ABC$ then we know that $$\tan \frac{C}{2} = \frac{r}{CD}$$ so that $$CD = \frac{r}{\tan(C/2)} = \dfrac{r}{\tan\left(\dfrac{\pi - \theta - \beta}{2}\right)} = r\tan\left(\frac{\theta + \beta}{2}\right)\tag{4}$$ Note further that the inradius $r$ is given by the formula \begin{align} r &= \frac{\Delta}{s}\notag\\ &= \frac{bc\sin 2\beta}{a + b + c}\notag\\ &= \dfrac{\dfrac{\sin^{2}\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}}{\dfrac{\sin\theta\{\sin(\theta + \beta) + \sin(\theta - \beta) + \sin 2\beta\}}{\sin(\theta + \beta)\sin(\theta - \beta)}}\notag\\ &= \frac{\sin\theta\sin 2\beta}{2\sin\theta\cos\beta + 2\sin\beta\cos\beta}\notag\\ &= \frac{2\sin\theta\sin\beta\cos\beta}{2\cos\beta\{\sin\theta + \sin\beta\}}\notag\\ &= \frac{\sin\theta\sin\beta}{\sin\theta + \sin\beta}\tag{5} \end{align} If $e$ is the eccentricity then we know that $$e = \frac{\text{distance of focus from center}}{\text{length of semi major axis}} = \frac{(a/2) - CD}{a/2} = \frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as \begin{align} e &= \dfrac{a - 2r\tan\left(\dfrac{\theta + \beta}{2}\right)}{a}\notag\\ &= 1 - \frac{2r}{a}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \frac{2\sin\theta\sin\beta}{\sin\theta + \sin\beta}\cdot\frac{\sin^{2}\theta - \sin^{2}\beta}{2\sin\theta\sin\beta\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \frac{\sin\theta - \sin\beta}{\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \dfrac{2\cos\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \dfrac{2\sin\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\notag\\ &= 1 - \frac{\cos \beta - \cos \theta}{\cos \beta}\notag\\ &= \frac{\cos \theta}{\cos \beta} = \frac{\cos\theta}{\cos(\alpha/2)}\tag{6} \end{align} Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < \beta = \alpha / 2 < \theta \leq \pi/2$. If $\beta = \alpha/2 = \theta$ then the section becomes a parabola and if $\beta = \alpha/2 > \theta$ then the section becomes a hyperbola. For the angles given in question we have $e = \cos 70^{\circ}/\cos 30^{\circ} = 0.39493084\ldots$.

Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = \cos\theta/\cos\beta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.

[Math] Questions on the relation of the axis of a cone to its conic sections

Converting comments to answer, as requested:

The Dandelin spheres answer question (1): a focus of a conic section is the point of tangency of its plane with one of those spheres. Clearly, the point on tangency lies on the cone axis if and only if the plane is perpendicular to that axis; therefore, the axis contains a focus in, and only in, the case of a circle. The axis point has some relation to the conic section, but not one as interesting or useful as a focus. (As for "Does the diameter of the base of the cone pass through the focus of the hyperbola?": note that the cone has no base. The thing extends, and expands, infinitely-far.)

This notion of points seeming to move "out to infinity and come back on the other side" is not uncommon in analytical geometry. (It's kinda what happens along asymptotes, when you think about it.) You might be interested in studying Projective Geometry, which adds a "line at infinity" to the standard Euclidean plane; this broader context helps unify all the conic sections into a single kind of curve that has various different relations to that line. (The parabola, in particular, has its "second vertex" on that line.)

Best Answer

Related Solutions

[Math] Evaluate the eccentricity of the elliptical section of a right circular cone

[Math] Questions on the relation of the axis of a cone to its conic sections

Related Question