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

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?

• If you can find the focus by construction it shouldn't be hard - from there it's just Pythagoras. – Dan Uznanski May 26 '19 at 15:40
• It is known $F$ should be along projection line from vertex $V$ but nothing further at this stage. – Narasimham May 26 '19 at 17:49
• Is Joachimsthal 's thm in differential geometry relevant here? – Narasimham May 26 '19 at 18:51

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

• 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 $F$ ? – Narasimham May 27 '19 at 7:56
• 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. – Intelligenti pauca May 27 '19 at 11:27
• Is it like trisection problem of an angle that is proved to be impossible by Euler? – Narasimham May 28 '19 at 7:45
• I don't see anything impossible in this case: you can easily construct all the above lengths. – Intelligenti pauca May 28 '19 at 11:11
• 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. – David K May 28 '19 at 19:38

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

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.