(X) ( a d/2 e/2) q(M) = (Y) × (d/2 b f/2) × (X, Y, Z) = 0 (Z) (e/2 f/2 c )All the proper conic sections are then "equivallent", by just changing the homogenous coordinate system.
All properties of pole/polar on a circle are immediately converted by replacing circle with conic section
(except those dealing with distances and orthogonality) :
If U ∈ polar of V, then V ∈ polar of U
If U is the pole of d and V the pole of d', then UV is the polar of the intersection point I=d∩d'
If A lies on Γ, the polar of A is the tangent in A.
Hence if A and B on Γ, the intersection point of the tangents in A and B is the pole of AB.
Some poles and polars in this figure :
AB polar of P (intersection of tangents in A and B)
d polar of U, construction of the polar :
two lines from U intersect Γ in A,B,C,D. I = AC∩BD, V = AD∩BC, the polar is IV.
UV polar of I
AC polar of T, d polar of U, d' polar of V, hence TUV in line on the polar of I = intersection of d, d' and AC
Reciprocal polar transform :
an application which transforms P ↔ polar of P.
|An homography between two pencils of line defines a conic section and reciprocally.|
Given a conic section Γ, two points A and B on Γ, O the pole of line AB (hence intersection of the tangents in A and B).
Let C any point on Γ other than A and B. There is one and only one homography h of A* → B*, with center O, which transforms AC into BC (with center O means it transforms AB of A* into BO of B*, and AO of A* into AB of B*)
For any point M ∈ Γ, the points O, I = AM∩BC and V = AC∩BM are inline on the polar of U = AB∩CM.
This just says that line IV goes through O, hence AM is transformed into BM by the homography h
The reciprocal is obtained from the only conic section tangent in A to OA, in B to OB and going through C = d∩h(d), then defining an homography h' as above, and prooving that h = h' (they are the same on 3 lines of A* → B*).
The draggable cyan, green and magenta points define 3 lines of A* and the corresponding lines of B*,
hence an homography A* → B*.
The center O of this homography is then constructed.
A current line (d) of the pencil A* is choosen by the red draggable point, then transformed by the homography centered in O into the image line in B*.
A* and B* may be defined from the intersection points of A* with a line d not going through
A and with a line d' not going through B.
The homography A* → B* = h(A*) is then equivallent to an homography between lines d and d', or even an homography of a line d onto itself, d going neither through A nor B.
This gives an equivallent theorem :
Given two points A and B and a line not going through A and B.
An application h of d → d is an homography iff
the locus of intersections of AM and Bh(M), M∈d is a conic section.
At last, an homography keeping the cross ratio :
|6 points ABCDEF are on a same conic section iff (AC,AD,AE,AF) = (BC,BD,BE,BF)|
An homography on d induces then an homography on Γ, keeping the cross ratio on Γ.
Such an homography can be extended to the whole plane into some homography keeping Γ (or conversedly, an homography of the plane keeping Γ can be restricted to Γ into an homography on Γ)
|Any homography of a conic section onto itself is the product of two involutions|
| An homography h of a conic section Γ on itself is an involution iff
for any point M on Γ the line Mh(M) goes through a fixed point S
The conic section is here defined by the 5 draggable red points.
The involution is definie by A→A', B→B', I draggable gives M→M'
Let S the intersection of AA' and BB', and d the polar of S.
S and d define an homology on the plane, A→A'
and as (S,U,A,A') = -1, it is an involution.
It exchanges (polar of S !) each point M with M', (S,I,M,M') = -1.
The restriction of this involution to Γ is then an involution on Γ.
| Let α an homography on a conic section.
For any pairs M,N → M',N', the intersection point MN'∩M'N is on a fixed line :
The homography axis.
Let's consider the homography α defined by A→A', B→B', C→C' on Γ.
In the applet, A,A',B,B',C define also the conic section itself
The yellow point defines C' hence the homography (ABC) → (A'B'C')
The cyan point defines a current point M.
The application tA transforming any line AM' of pencil A* into line A'M of pencil A'*
keeps the cross ratio of lines,
as the cross ratio A'(M1,M2,M3,M4) = A'(M'1,M'2,M'3,M'4) = A(M'1,M'2,M'3,M'4) :
the homography α on Γ keeps the cross ratio, which doesn't depend on
the choice of A or A' on the conic section.
The application tA is then an homography (between pencils of lines).
tA(AA') = A'A hence this homography is a projection. Therefore AM'∩A'M is on a fixed line : the axis of this projection dA.
Note : properties of projection between pencil of lines are deduced by duality from properties of projections between two lines.
Similarily, the homography tB B* → B'* gives BM'∩B'M ∈ dB.
We have just to prove that dA = dB
Point I = AB'∩A'B is common to these two lines.
Let J = A'C∩AC', of course on dA
Consider the involution β from center J transforming B into D1 and B' into D. Let D' = α(D).
α and β keeping the cross ratio : (A,B,C,D) = (A',B',C',D') by α and (A,B,C,D) = (C',D1,A',B') by β
As exchanging points in cross ratio (C',D1,A',B') = (A',B',C',D1) keeps it unchanged, (A',B',C',D') = (A',B',C',D1) hence D1 = D'
In other words J = BD'∩B'D hence J ∈ dB
Therefore dA = dB = IJ
|The opposite sides of an hexagon inscribed in a conic section intersect into 3 in line points
An elementary proof on a circle here.
By projection, this proof is extended to any proper conic section.
A direct projective proof uses an homography and the homography axis : A→D, E→B, C→F defines an homography on the conic section.
The intersect points AB∩DE, CD∩AF and BC∩EF are then in line on the axis of this homography.
the hexagon may be crossing : choosing arbitrarily the order of points ABCDEF, the "opposite sides" are (AB - DE), (BC - EF) and (CD - AF).
The reciprocal allows to construct points of a conic section defined by 5 points ABCDE.
Let M to be constructed, on any line Am through A, ABCDEM is an hexagon.
To construct M :
Lines AB and DE intersect in I
Any line Am intersects CD in J
BC intersects IJ in K
KE intersects Am in M.
Construction of a tangent in one of the given points results from the specific case when M comes into E :
Line Am = AE intersects CD in J, BC intersects IJ in K, KE is the tangent in E.
The applet allows to drag points ABCDE and point m, direction of the current line Am.
The 'nearly degenerate" cases (3 given points are in line) result into spurious drawings.
Conversedly, A,B can be the same point, given the tangent in A as line AB.
This allows to construct points of a conic section defined by 4 given points and the tangent in one of them, or 3 points and the tangents in two of them.
At last some given points might be at infinity, that is defined by their direction (we then draw parallels).
|The midpoints of intersections with parallel lines are on a same line.|
The polar Δ of D∞ is then called a conjugate diameter to direction d.
The reason of the word "diameter" is that all conjugate diameters to various directions d intersect in a same point : the center of the conic section. And also :
|The center is the pole of line at infinity|
We can then speak of conjugate diameters, considering Δ and the conjugate diameter
Δ' to direction Δ, Δ' // d.
When the two conjugate diameters are perpendicular, they are the axes of the conic section.
The tangents in the endpoints of the conjugate diameter to d are parallel to d.
Note : The polar Δ of D∞ goes through D'∞, hence the points D∞ and D'∞ (at infinity) are harmonic conjugates to Γ.