Intersection of two conic sections

The intersections of two conic sections are generally not constructible with compass and straightedge, as resulting into a general 4th degree equation.
The following cases are constructible, the 4th degree equation being reduced to constructible quadratic equations.

Same center

The problem is just to find the degenerated conic sections in the pencil defined by the two given conic sections.
We shall define here the two given conic sections by their focus F and F' and a circular directrix (F),
The draggable control points in the applet are really the center O, the apex S and focus F.
The circular directrix and the other focus are then deduced from elementary constructions.

Recall first a few basic constructions.

Construction of a current point

Let m a current point on the circular directrix (F).
Line Fm and the perpendicular bisector of F'm intersect in M on the conic section.
Even more, the perpendicular bisector of F'm is the tangent in M to the conic section.
Recall here the important property : M is the center of a circle going through F' and tangent to the circular directrix (F).
This property might be used as a definition of the conic section : locus of the center of circles etc.

Construction of intersection with any line (d)

In the applet, the line goes through O, but the construction works for any line.
This means to search on the line the center of a circle through F' and tangent to (F)
It then goes also through the reflection F" of F' through this line.
The problem is then reduced to find a circle going through F' and F" and tangent to (F), Apollonius problem 'PPC'.
Draw any circle through F' and F", hence centered on (d), and intersecting (F) in any A and B. Lines F'F" and AB intersect in I.
Let IT a tangent to (F) from I. The searched circle is the circumcircle to F'F"T, its center is then the intersection of (d) and the perpendicular bisector of F'T, or even simpler the intersection of (d) with FT.
The choice of "any" circle is here done manually, the green point d used to define the given line is used as center of this circle, to be chosen "outside" the conic section so that this circle intersects the circular directrix.
The construction fails if (d) is line FF', but then, constructing the vertices is much simpler !
The applet doesn't try to construct this specific case through a conditional construction.

Conjugate diameter of a given diameter (d)

Just apply the characteristic property : draw any secant parallel to (d) intersecting the conic section in M and N,
The conjugate diameter goes through the midpoint of MN.
Here M is the current point previously constructed, and the second intersection of the parallel to (d) through M is constructed as follows :
The tangent to circular directrix in m intersects F'F" in P, the second tangent Pn from P gives N as before.
Of course we don't construct this second tangent as a tangent, but directly Fn as reflection of Fm through FP, and N is intersection of this line with the parallel to (d) through M.

Sorry, your browser is not Java compliant
The current point is defined by the draggable cyan point m on the circular directrix.
The given line is defined by the green draggable point d.

Knowing these basic constructions, that we won't detail anymore, let's search the intersection of our two given conic sections.
As they have the same center, from symmetry, the intersection points build a parallelogram (for the diagonals intersect in their midpoint O)
The sides of this parallelogram are conjugate directions for both the conic sections.
Let (d) any diameter and (d') the conjugate diameter of (d) with respect to the first conic section Γ, then (d") the conjugate diameter of (d') with respect to the second conic section Γb.
We thus define (step 1) a projective homography (d) → (d") in the pencil of lines O* : green line (d) → orange line (d")
The directions of the parallelogram sides are then the fixed directions in this homography.
They are constructed by projecting (d) → (d") onto any circle going through O, the homography axis intersects the circle in fixed points.
For doing that (step 2) we chose any three points a,d,c and their transformed a",d",c", and we use the axis definition :
locus of the intersection points of ad" and a"d.
The chosen points are (a) focal axis of Γ, the conjugate is the perpendicular axis and (a") is the conjugate of this perpendicular axis with respect to Γb
We do the same with (c) the perpendicular axis of Γ and (c") the conjugate of the focal axis of Γ with respect to Γb
Finally (d) and (d") are a current direction, defined by draggable point d.

Sorry, your browser is not Java compliant

We thus get the directions (magenta) Ou, Ov of the parallelogram sides.
But we have now to construct the sides themselves.

Let's chose (step 3) any line (L) intersecting the conic sections in (M, M') and (Mb, M'b)
The pencil of conic sections defined by Γ and Γb defines on (L) a Desargues involution fully determined by M ↔ M' and Mb ↔ M'b
If this involution is hyperbolic, it has two real fixed points I and J.
For that, just chose (L) so that one chord of MM' or MbM'b is strictly inside the other.
Also to get an effective construction, these fixed points shouldn't be at infinity, hence (L) shouldn't be parallel to the previously constructed conjugate diameters. This allows however a wide choice for (L).
In the applet (L) is defined by the two "current" points M and Mb of the two conic sections (driven by the cyan points on the circular directices).
Let's construct then (step 4) the fixed points of this involution, that we obtain by projecting it on any circle, for instance the one previously chosen.
The axis of the involution on the circle is constructed : line through the intersection points of mmb with m'm'b, and of mm'b with m'mb
The intersections of this axis with the circle give the fixed points I and J after projecting back to (L).
The involution M ↔ M' on (L) is then characterized by (M,M',I,J) = -1, harmonic range.
Let Q the intersection of (L) with diameter Ou and E,E' the intersections of (L) with the searched degenerated conic section, made of the two common chords parallel to Ou.
Q is obviously the midpoint of EE' and we can then construct (step 5) E and E' through QE² = QE'² = QI.QJ (classical).
We get then the common chords as the parallels to Ou through E and E', then the intersections of these lines with Γ ( or with the other degenerated conic section parallel to Ov, constructed in a similar way).

Important note :
The choice of the "any points" impacts the construction.
If we change the conic sections,we might also sometimes change (manually) these construction points ("ctrl pnt" button) :

Current points on the conic sections : m,M and mb,M. it is the most critical condition.
Any line (d)
Any circle : center ω (w), not so much important

The construction even completely fails if the two conic sections have the same axes :
We should then chose a second arbitrary line in step 1.
However in this case the construction is simpler : the fixed directions are just the axes !
An automatic choice of these control points would lead to an irrelevant figure, most of the points being outside the drawing and would make the construction even more complicated (165 constructed geometric elements in the applet)
Even the simple construction of an intersection or the conjugate requires several intermediate elements as shown in the preliminary constructions.
Even just project a direction (d) on a circle requires two elements ! Just intersecting (d) with the circle doesn't work : we must grant that it is the other intersection point than O etc.
Although it is obvious manually, it is not so simple inside an applet, and it is often needed to use artificial constructions, for instance in this case it is the reflection of O through the perpendicular to (d) from center ω.

One common focus

The conic section being defined as the locus of the centers of circles going through this common focus and tangent to the circular directrix (or the directrix for parabolas), the problem is reduced to Apollonius problem in the case PCC : one point (the common focus) and two circles (the two circular directrices) or PLC (a parabola, the directrix) or PLL (two parabolas, the directrices).

A common axis

Analytically, the 4th degree equation reduces to just a quadratic, hence constructible.
In the previous method, there are always 0 or 4 real intersection points (or two double points if the conic sections are tangent)
Here, there might be just two real points.
The pencil defined by the two conic sections might then have two real base points and two imaginary !
Also the conic sections may as well be parabola(s)

Here also, the key is to construct a degenerated conic section in the pencil, and the Desargues involution.

Details (these applets are too big to fit in the same page)

Known points

If two of the intersection points are already known, the two others can be constructed.
The construction might even be done with straightedge alone if three intersection points are known, or if one conic section is a circle.
This construction has been seen here and there


Home Arithmetic Geometric Misc Topics Scripts Games Exercices Mail Version Française Previous topic Next topic