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As we saw earlier, a conic section is the intersection of a (double-ended) right circular cone with a plane.

**Remarks**

- As we saw before, all conic sections are projectively equivalent to a circle.
- An alternative definition of a conic in
**R***P*^{2}is to take the set of points satisfying an equation of degree 2.

That is, in homogeneous coordinates:

*ax*^{2}+*by*^{2}+*cz*^{2}+ 2*fyz*+ 2*gxz*+ 2*hxy*= 0 or*x*^{t}*A*= 0 where*x**A*= and= .*x*

This however includes some singular conics like:

or*x*^{2}+*y*^{2}+*z*^{2}= 0 ⇒ empty set

or*x*^{2}-*y*^{2}= 0 ⇒ two lines*x*=*y*and*x*= -*y*

or*x*^{2}= 0 ⇒ one line*x*= 0- In
**C***P*^{2}the non-singular conics correspond to det(*A*) ≠ 0- One can prove the equivalence of non-singular conics using the usual diagonalisation arguments.

Some of the standard Euclidean results about circles now give results about conics.

Here is a result proved by the French mathematician Michel Chasles (1793 to 1880) in 1852.

**Chasles' theorem**

*Let A, B, C, D be distinct points on a *(*non-singular*)* conic. If P is another point on the conic then the cross-ratio of the pencil PA, PB, PC, PD does not depend on the point P*

**Proof**

By Exercises 8 Question 4, the cross-ratio of a pencil is the cross-ratio of the four points at which the pencil meets any line.

Project the conic into a circle. The cross ratio of the pencil is determined by the angles beween the lines *PA*, *PB*, *PC*, *PD* and if we move the point on the circle these angles are unchanged.

**Remark**

From this last result, if four points lie on a conic we may unambiguously talk about their cross-ratio.

We can now prove a result discovered by the French mathematician Blaise Pascal (1623 to 1662) when he was 16 years old. He called it the *mysterium hexagrammicum* = mystic hexagram.

**Pascal's theorem**

*If a hexagon is inscribed in a *(*non-singular*)* conic then the meets of opposite sides are collinear.*

**Proof**

Project from 1: (2 , 4 ; 5 , 6) = (*Q* , 4 ; 5 , *X*)

Project from 3: (2 , 4 ; 5 , 6) = (*P* , *Y* ; 5 , 6)

Now consider the two pencils with vertex *R* and we get

(*RQ* , *R*4 ; *R*5 , *RX*) = (*RP* , *RY* ; *R*5 , *R*6)

But *R*6 = *RX*, *R*4 = *RY* and so *RP* = *RQ* and *P*, *Q*, *R* are collinear.

**Remarks**

- If the conic is singular this result is
*Pappus's theorem*(named after Pappus of Alexandria (about 300 AD). See Exercises 8 Question 5. - If one lets points on the conic coincide, their joins become tangents to the conic and one can deduce various new theorems. It is sometimes claimed that Pascal deduced as many as 400 corollaries from his theorem.
- If one is prepared to use
*PGL*(3,**C**) rather than*PGL*(3,**R**) then one can project any conic into a circle and any line into any other line. So projecting the conic into a circle and*PQ*into the line at infinity we get a diagram as shown on the right. In this 12 and 45 are parallel and so too are 23, 56. One may then use angles to show that 16 and 34 are parallel and the result follows.

The dual of a set of points lying on a conic (a locus) is a set of lines being tangent to a conic (an envelope). We can then dualise Pascal's theorem to get a result proved by the French mathematician Charles-Julien Brianchon (1783 to 1864).

**Brianchon's theorem**

*If a hexagon is circumscribed to a conic then the joins of opposite points of the hexagon are concurrent.*

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