Conics on the real projective plane

In this section, we undertake the study of nondegenerate conics on ${\Bbb R}P^2$. Our aim is to relate the algebraic approach of the previous sections to the classical synthetic approach of Chasles and Steiner.

The first important result about conics is that, up to projective transformations, they are all the same. More precisely:

Exercise 4.1 (05)   Show that if $Q$ and $Q'$ are two quadratic forms in ${\Bbb R}^3$ that define two nondegenerate (and non-empty) conics in ${\Bbb R}P^2$, then there exists an invertible linear transformation $M$ such that either $Q' = Q \circ M$ or $Q' = - Q \circ M$. Conclude that all nondegenerate conics on ${\Bbb R}P^2$ are projectively equivalent.

Remark. From now on all conics will be nondegenerate and nonempty.

As a consequence of the previous exercise, we have that it is possible to define the cross-ratio of four points on a conic.

Exercise 4.2 (10)   Let $A,B,C,D$ be four distinct points on a conic ${\cal C}$. If $E$ and $F$ are any two points on ${\cal C}$ that are different from the previous four points, the cross-ratios $[EA,EB,EC,ED]$ and $[FA,FB,FC,FD]$ are equal (Fig. 2). Use this to define the cross-ratio of four points on a conic. Hint: assume ${\cal C}$ is a circle and then use the conclusion of exercise 4.1 to reduce the problem to this case.

Figure 2.

By duality, we may also define the cross-ratio of four lines tangent to a conic: if $a,b,c$, and $d$ are four lines tangent to a conic, take a fifth line $e$ tangent to the same conic and define the cross-ratio $[a,b,c,d]$ as the cross-ratio of the points $e \cdot a, e \cdot b, e \cdot c$, and $e \cdot d$ on the line $e$ (Fig. 3)

Figure 3.

Exercise 4.3   If $X$ and $Y$ are two points on a conic ${\cal C}$, we may set up a correspondence between the pencil of lines passing through $X$ and the pencil of lines passing through $Y$: if $\ell$ is a line passing through $X$, then it either intersects the conic at some other point $Z$ or it is the tangent line to ${\cal C}$ passing through $X$. In the first case, we associate to $\ell$ the line $YZ$; in the second, the line $YX$ (Fig. 4). Use exercise 4.2 to show that this correspondence preserves cross-ratios and is, therefore, a projective transformation between the pencils.

Figure 4.

Conversely, we have the following beautiful construction of conics due to Steiner.

Theorem 4.1   Let $f$ be a projective transformation between the pencil of lines passing through a point $X$ and the pencil of lines passing through a point $Y$. The set

$${\cal C} :=\{\ell \cdot f(\ell) :$$    $\ell$ is a line passing through $X$$$\}$$

is a conic.

Exercise 4.4 (10)   Show that if the projective transformation is a perspectivity, then the construction yields a degenerate conic.

If $T$ is a projective transformation taking $X$ to $[0:1:0]$ and $Y$ to $[1:0:0]$, then the pencils of lines passing through $X$ and $Y$ are taken to the pencils of vertical and horizontal lines on the plane $z = 1$ (Fig. 5). Furthermore, ${\cal C}$ is transformed into the graph of a function $g$ from the $x$-axis to the $y$-axis.

Figure 5.

Exercise 4.5 (10)   Use that $T \circ f \circ T^{-1}$ is a projective transformation between the pencil of lines passing through $[0:1:0]$ (the pencil of vertical lines on the plane $z = 1$) to the pencil of lines passing through $[1:0:0]$ (the pencil of horizontal lines) to show that $g$ is of the form $g(x) = (ax + b)/(cx + d)$. Conclude that $T({\cal C})$ is a hyperbola on the plane $z = 1$ and use this to prove theorem 4.1.

Exercise 4.6 (10)   State the dual of Steiner's theorem. This is Chasles' theorem.

Using Steiner's theorem, we can easily prove that five points determine a conic. Indeed, let $A,B,C,D$, and $E$ be five points on the plane. If three or more of these points are collinear then it is easy to see that they all lie on a degenerate conic. Assume now that no three of these points are collinear. Remark that there is a unique projective transformation $f$ between the pencil of lines passing through $A$ and the pencil of lines passing through $B$ such that $f(AC) = BC$, $f(AD) = BD$, and $f(AE) = BE$ (Fig. 6). By Steiner's theorem $f$ defines a conic that passes through all five points.

Figure 6.

By duality, we have that given five lines, there is a unique conic that is tangent to all of them.

Since five points determine a conic, it is clear that six points are not on the same conic unless they satisfy some special condition. This condition, discovered by the mathematician and philosopher Blaise Pascal, is one of the earliest and prettiest results on projective geometry.

Theorem 4.2 (Pascal's theorem)   The six vertices of a hexagon lie on a conic if and only if the three points obtained by intersecting the three pairs of opposite sides are collinear (Fig. 7).

Figure 7.

The following two exercises prove that if the six vertices of a hexagon lie on a conic the three points obtained by intersecting the three pairs of opposite sides are collinear. The proof of the converse uses the same ideas and is left for the reader as a less structured exercise.

Consider the transformation $f$ that takes the points of the line $ED$ to the points of the line $EF$ that is defined by the following construction:

1. If $X \in ED$, draw the line $AX$.
2. Let $A'$ be the second point of intersection of the line $AX$ and the conic.
3. Draw the line $CA'$.
4. Let $Y := f(X)$ be the point $EF \cdot CA'$.

Exercise 4.7 (05)   Prove that the transformation $f$ is a perspectivity.

Exercise 4.8 (05)   Show that $f(P) = R$ and that the center of perspective is $Q$. Verify that this proves the first part of Pascal's theorem.

Exercise 4.9 (10)   Show that if the three points obtained by intersecting the three pairs of opposite sides of a hexagon are collinear, its vertices lie on a conic.

The dual of Pascal's theorem is the following result originally dicovered by Brianchon.

Theorem 4.3 (Brianchon's theorem)   The six sides of a hexagon are tangent to a conic if and only if the three lines obtained by joining the three pairs of opposite vertices are concurrent (Fig. 8).

Figure 8.