3
$\begingroup$

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 $q$.

Conic section sketch

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?

Finding Foci Hyperbola Intersection

$\endgroup$
3
  • $\begingroup$ If you can find the focus by construction it shouldn't be hard - from there it's just Pythagoras. $\endgroup$ May 26, 2019 at 15:40
  • $\begingroup$ It is known $F$ should be along projection line from vertex $V$ but nothing further at this stage. $\endgroup$
    – Narasimham
    May 26, 2019 at 17:49
  • $\begingroup$ Is Joachimsthal 's thm in differential geometry relevant here? $\endgroup$
    – Narasimham
    May 26, 2019 at 18:51

2 Answers 2

3
$\begingroup$

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.

enter image description here

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 $c={1\over2}\sqrt{VW^2−VV'\cdot WW'}$.

enter image description here

$\endgroup$
12
  • $\begingroup$ Can the focus be now marked from above definition with a point here or in any other projection/view? Is the intersection of axis of symmetry and (red) cutting plane projection the focus $F$ ? $\endgroup$
    – Narasimham
    May 27, 2019 at 7:56
  • $\begingroup$ By the same formulas given in the linked answer we can find the distance between a focus and the center of the ellipse: $$c={1\over2}\sqrt{VW^2-VV'\cdot WW'}.$$ The focus is not, in general, the intersection between the plane and the axis of the cone. $\endgroup$ May 27, 2019 at 11:27
  • $\begingroup$ Is it like trisection problem of an angle that is proved to be impossible by Euler? $\endgroup$
    – Narasimham
    May 28, 2019 at 7:45
  • $\begingroup$ I don't see anything impossible in this case: you can easily construct all the above lengths. $\endgroup$ May 28, 2019 at 11:11
  • 1
    $\begingroup$ The foci can be located by Dandelin spheres. In your figure you can construct a circle with center on the cone’s axis, tangent to the sides of the cone and to the cutting plane. The point of tangency on the cutting plane is a focus. $\endgroup$
    – David K
    May 28, 2019 at 19:38
1
$\begingroup$

Here is a geometric construction of the foci and latus rectum of an ellipse based on the Dandelin spheres:

enter image description here

Given a cone with vertex at $V$ and semi-vertex angle $\gamma$ cut by a plane at angle $\alpha$ to the axis of symmetry, in a plane through the axis of symmetry perpendicular to the cutting plane the cutting plane intersects the cone at points $A$ and $B$ so that $AB$ is the major axis of the ellipse. Bisect angle $\angle VAB$ and let $P$ be the intersection of the angle bisector with the axis of symmetry. Drop a perpendicular from $P$ to $AB$ intersecting $AB$ at $F.$ Then $F$ is a focus of the ellipse and $PF$ is a radius of the corresponding Dandelin sphere.

The other focus is similarly constructed by bisecting the angle $\angle B$ as shown in the figure (the other bisector would give us $P$ again), finding the intersection $Q$ of the bisector and the axis of symmetry, and dropping a perpendicular from $Q$ to $F'$ on $AB$; then $F'$ is the other focus and $QF'$ is a radius of its Dandelin sphere.

A plane through $F$ perpendicular to the axis of symmetry intersects the cone in a circle perpendicular to the plane of the figure, with diameter $CD$ in the plane of the figure. The semicircle $CD$ in the plane of the figure is congruent to one half of that circle. Construct a line through $F$ perpendicular to the diameter $CD$ intersecting the semicircle at $G.$ Then $FG$ is the semi-latus rectum.

There is a corresponding construction for the foci and semi-latus rectum of a hyperbola. The construction for the parabola has only one sphere and one focus.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .