Projective conic sections - constructions

The constructions here are mainly about conic sections defined by 5 points (or through duality by 5 tangents), including the cases when some points overlap.
The construction of a current point ("drawing") of the conic section from 5 points has been used here many times, it uses the Pascal theorem. We won't detail again, see here.

In case of overlapping points, the "line" joining two overlapping points is the tangent in that point. This allows to adapt the previous construction into :
- Construct the tangent in one of the given points (or a current point)
- Draw the conic section given one tangent, the contact point and 3 other points
- or given two tangents and their contact points, and another point

Intersections with a line

When the line goes through one of the given points, the previous construction gives at once the second intersection point.
The question is more difficult when it is any line.

Sorry, your browser is not Java compliant. A first method : transform the conic section into a circle by a projective homology.
The line is transformed into a line that intersects the circle in the images of the searched intersections.

Construct the tangent in one of the given points O, then draw any circle tangent in O, which is then the transform of the conic section in an homology of pole O.
The transforms of points P1, P2, P3 → p1,p2,p3 are constructed just by intersecting the circle with lines through O : OP1 ...
The transformed of the given line (d) is constructed by drawing the transformed of two of its points : here U and V, U intersection of P1P2 with (d) hence u intersection of p1p2 with OU, and similarily for V on P2P3.
The transformed of (d) is then line uv, whose intersections x, y with the circle are sent back on (d) into X, Y by the rays Ox and Oy

In the applet, the red draggable points define the conic section, the blue points define the line (d).
The yellow point chooses the tangent circle.

Other possible constructions appear finally less simple (need to construct more points)
For instance :
The pencil of conic sections through ABCD intersects (d) and define on (d) a Desargues involution I, exchanging the intersection points on (d).
Similarily, the pencil defined by ABCE defines an involution J.
The searched intersection points X, Y correspond both in I and J, hence are fixed points in the homography I◦J.
These fixed points are constructed by projecting this homography of (d) onto a circle. The fixed points on the circle are the intersections if any with the homography axis.

Sorry, your browser is not Java compliant.
Each involutions on (d) is defined by the intersections of the degenerate conic sections of the two pencils.
As they are involutions, two points suffice. (AB,CD) and (AD,BC) give M1→M'1, M2→M'2
We can spare a few lines by choosing for the second pencil (AB, CE) and (AE, BC), that is M1=N1→N'1, N2→N'2=M'2
Each involution is projected on a circle, the centers of these involutions are constructed : intersections I of m1m'1 with m2m'2 and J of n1n'1 with n2n'2.
The product of these involutions is an homography with axis IJ, the fixed points are intersections x and y of IJ with the circle, then projected back on line (d) into X and Y.
In the applet the conic section is defined by the draggable red points ABCDE, the line is defined by the blue points. The circle and the projection are freely choosen by the draggable yellow points.

At last a third construction :
Recall the polar of a point is the locus of all harmonic conjugates, real or imaginary, of this point. That is the harmonic conjugates of P with respect to the intersection points of any secant through P.

Consider then the polar of point at infinity on (d).
It is the "conjugate diameter" of direction (d), locus of midpoints of secants DD' parallel to (d), as (∞, m, D', D) = -1 ⇔ m midpoint of DD'
This polar intersects then (d) in point M, midpoint of XY.

Consider now the polar of any other point P of (d).
Different from M, otherwise we won't get a new relation, but the same : M midpoint of XY.
Recall the classical construction of the polar of P :
Draw any two secants PCC' and PDD'.
CD and C'D' intersect in I, CD' and C'D intesrect in J
IJ is the polar of P.

Let U the intersection of the polar of P with (d). (P,U,X,Y) = -1 and choosing a coordinate system at midpoint M : MX² = MY² = MP.MU
The construction of MX = MY is then a classical : the length of the tangent MT from M to circle with diameter PU.
If M is inside the segment PU, there is no real tangent, neither real intersection points.

Sorry, your browser is not Java compliant.
Of course the parallels to (d) and the secants from P are choosen as going through one of the given points, otherwise it is a vicious circle !
Their second intersection (C' D' C" D") is then constructed from Pascal hexagon.
In the above applet, the 5 red points including C and D define the conic section, the two blue points including P define the line (d)

Intersection of two conic sections

Algebraically, this intersection is of degree 4, hence can't be constructed with compass and straightedge in general case.
Reducing the degree happens if we already know two of the intersection points.
That is, given two conic sections intersecting in A and B, construct the two other intersection points.
Theorem :
Let two conic sections intersecting in A,B,C,D.
Any line through A intersects the conic sections again in M,M'
Any line through B intersects the conic sections again in N,N'

MN, M'N' and CD concur in a same point.

Let's apply the Pascal theorem to the hexagons MACDBN and M'ACDBN'
On Γ, the intersection points I = MA∩DB, J = AC∩BN and O = CD∩MN are in line,
hence O is intersection of IJ and CD.
On Γ', the intersection points I' = M'A∩DB, J' = AC∩BN' and O' = CD∩M'N' are in line,
hence O' is intersection of I'J' and CD.
As line MA is line M'A, I = I', and similarily J = J', hence O = O'.

By repeating the construction, we get two points O1 and O2, which define then line CD, hence the intersection points C and D.
In the applet, the two conic sections are defined by the 5 red points, including the common points A and B.
Points MM'NN' are easily constructed, as lines MM' or NN' go through one of the given points.
The cyan points choose the lines AM1M'1, AM2M'2, BN1N'1, BN2N'2.
But the intersection with line CD is more difficult (intersection with any line, see above)
Here in yellow, the conic section Γ is transformed into a circle tangent in B, images o1, o2 of points O1 and O2 is constructed.
Intersection of image line o1o2 with the circle is sent back on line O1O2 into C and D.
Sorry, your browser is not Java compliant.
Using an homology to construct C and D is useless, and the construction is simplified, if 3 of the intersection points are given (ABC given, to construct D), or if one of the conic sections is a circle.
The construction is then done with straightedge alone.

: conic section defined by...

 

 

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