PROJECTIVE GEOMETRY COURSE

*Chapter 4*: PROJECTIVE PROPERTIES OF CONICS

§ 20: *Classification of quadratic surfaces*

A quadratic surface is the set of points in ℜ^{3} that satisfy an equation of the form

a_{1 1}x_{1}^{2} + a_{2 2}x_{2}^{2} + a_{3 3}x_{3}^{2} + 2 a_{1 2}x_{1}x_{2}
+ 2 a_{1 3}x_{1}x_{3} + 2 a_{2 3}x_{2}x_{3} = 0 (for given real numbers a_{i j}).

Such a quadratic surface exists of straight lines through O, and hence we call it a cone with top O. In this section we will see that the intersection of the quadratic surface with the plane
x_{3} = 1 is a parabola, a hyperbola or an ellipse.

Speaking in projective terms, each straight line through O is a point, and the quadratic surface a *conic* (parabolas, ellipses and hyperbolas are (together with their degenerated versions)
exactly the possible intersection figures of of a common right circular cone and a plane).

The inhomogeneous equation is a_{1 1}x_{1}^{2} + a_{2 2}x_{2}^{2} + a_{3 3} + 2 a_{1 2}x_{1}x_{2}
+ 2 a_{1 3}x_{1} + 2 a_{2 3}x_{2} = 0, or (in matrix notation)

Short notation of this matrix equation: x^{t} A x + 2 a^{t} x + a_{3 3} = 0.

The symmetric matrix A has two real eigenvalues λ_{1} and λ_{2} (that possibly coincide), and there is in ℜ^{2} an orthonormal base consisting of
eigenvectors of this matrix: say r_{1}, r_{2}, where Ar_{1} = λ_{1}r_{1}, Ar_{2} = λ_{2}r_{2}.

Use coordinates y_{1} and y_{2} with respect to this new base, so x = y_{1}r_{1} + y_{2}r_{2}, or, in an other notation, x = Ry where
r_{1} stands in the first column of R and r_{2} in the second column. Then the matrix R is orthogonal, which means R^{t} = R^{-1}. Now check by calculation
that R^{t}AR is the diagonal matrix with eigenvalues λ_{1} and λ_{2} on the diagonal.

So in the new coordinates the matrix equation becomes y^{t}R^{t} A Ry + 2 a^{t} Ry + a_{3 3} = 0, or

short λ_{1}y_{1}^{2} + λ_{2}y_{2}^{2} + 2 b_{1}y_{1} + 2 b_{2}y_{2} + c = 0.

If λ_{1}λ_{2} = 0 we get the equation of a parabola.

Else, if λ_{1}λ_{2} is positive, we have the equation of an ellipse, and if λ_{1}λ_{2} is negative the equation of a hyperbola.

*O79* Elaborate these calculations with x_{1}^{2} + 2 x_{1} x_{2} + x_{2}^{2} + 3 x_{1} + 1 = 0, and with
2 x_{1}^{2} + 2√2 x_{1} x_{2} + 3 x_{2}^{2} + x_{2} = 0.

Now we are going to find the equation of the tangent in a point of the conic.

Let x^{t}Ax = 0 be the equation (in homogeneous coordinates) of a conic, and let λp be a point of the conic, so that p^{t}Ap = 0.

Let λp + μq represent an arbitrary straight line through λp.

Intersection with x^{t}Ax = 0 gives λ^{2} p^{t}Ap + λμ(p^{t}Aq + q^{t}Ap) + μ^{2} q^{t}Aq = 0, so also
μ(2λ p^{t}Aq + μ q^{t}Aq) = 0.

Now μ = 0 is a *double* solution of this equation (and hence λp a double intersection point of the line and the conic) if and only if p^{t}Aq = 0.

If x = λp + μq, then p^{t}Aq = 0 ↔ p^{t}Ax = 0, so p^{t}Ax = 0 is the equation of the tangent in λp to the conic x^{t}Ax = 0.

So we find the equation of the tangent in a point of the conic with the method of 'honest distribution'.

*O80* Which equation has the tangent in λ(0,-1,1) to the conic x_{1}^{2} + 2 x_{2} x_{3} + x_{2}^{2} +
3 x_{1} x_{3} + x_{3}^{2} = 0?