Perhaps, this

https://en.wikipedia.org/wiki/Algebraic_curve#Rational_curves

will help. In Russian wikipedia the reference is

Yu. I. Manin, “Rational points on algebraic curves over function fields”, Amer Math. Soc. Transl. (2), 50 (1966), 189–234

Hello, thank you Alexei,

I will study the relationship with the unicursal curves.

BR

Gabriel

The General Situation

Let k denote any field, and let K be any extension field of k, possibly K = k.

A line defined over k is an equation

L : Ax + By = C, A, B, C ∈ k, where at least one of A, B is nonzero. A K-rational point of L is a solution

(x, y) ∈ K2 of L. The set of K-rational points of L is denoted LK.

Similarly, a conic curve defined over k is an equation C : ax2 + bxy + cy2 + dx + ey + f = 0, a, b, c, d, e, f ∈ k, where at least one of a, b, c nonzero. A K-rational point of L is a solution (x, y) ∈ K2 of C, and the set of K-rational points of L is denoted CK. (Note: CK may not contain any points at all. For example, let k = K = R and consider the conic curve C : x2 + y2 = −1.)

Proposition 1.1. Suppose that CK contains a point P = (xP , yP ) not in LK.

Then the points of CK other than P are in bijective correspondence with the points

of LK.

Proof. First note that after a coordinate translation, we may let

P = (0, 0), although now the coefficients of L and C could lie in K rather than k.

For any given point Q = (xQ, yQ) ∈ CK such that Q 6= P, let t = yQ/xQ ∈ K

and then solve the equation L(x, tx) for a unique xR ∈ K. Let yR = txR. The

point R = (xR, yR) ∈ LK is collinear with P and Q.

Conversely, for any given point R = (xR, yR) ∈ LK such that R 6= P, let

t = yR/xR ∈ K and then consider the equation C(x, tx). This quadratic equation

has x = 0 as a solution, but after dividing the equation through by x there is a unique second solution xQ ∈ K. (Possibly xQ = 0 as well.) Let yQ = txQ. The point Q = (xQ, yQ) ∈ CK is collinear with P and R.

The argument here has left out the case where all the x-coordinates agree. This situation can be handled as a special case. 

Note that the fields k and K in this discussion are completely general. For example, k could be the field of p elements for some prime p, and K could be the field of q = pe elements for some positive integer e.

2. The Circle in Particular

Now define

L : x = 0,

C : x2 + y2 = 1,and let P = (−1, 0), an element of Ck for any field k.

Given a point Q = (xQ, yQ) ∈ CK, the corresponding point on LK is

R =0,yQxQ + 1

Conversely, given a point R = (0, yR) ∈ LK, let t = yR. We seek a point Q =

(x, t(x + 1)) ∈ CK.

But

x2 + y2 = ((x + 1) − 1)2 + t2(x + 1)2 = (1 + t2)(x + 1)2 − 2(x + 1) + 1,

so we want

(1 + t2)(x + 1)2 − 2(x + 1) = 0,

or (1 + t2)(x + 1) = 2, or x + 1 = 2/(1 + t2). Since y = t(x + 1) it follows that

y = 2t/(1 + t2), so that finally,

Q =1 − t21 + t2,2t1 + t2



3. An Application From Calculus

Let θ denote the angle to a point (x, y) ∈ CR. Then the quantity t in the previous discussion is

t = tan(θ/2).

Thus θ = 2 arctan(t), giving the third of the equalities

cos(θ) = 1 − t2

1 + t2, sin(θ) = 2t

1 + t2, dθ =2 dt1 + t2

The rational parametrization of the circle gives rise to the substitution in elementary calculus that reduces any integral of a rational function of the transcendental

functions cos(θ) and sin(θ) of the variable of integration θ to the integral of a rational function of the variable of integration t.

4. An Application From Elementary Number Theory

A primitive Pythagorean triple takes the form (a, b, c) ∈ Z3, a2 + b2 = c2, a, b, c > 0, gcd(a, b, c) = 1.

It follows that in fact a, b, and c are pairwise coprime. We may take a odd, b even, and c odd. (Inspection modulo 4 shows that the case where a and b are odd but c is even can not arise.)

Given such a triple, let x = a/c and y = b/c. Then (x, y) is a point of CQ,

(x, y) = 1 − t2

1 + t2,2t1 + t2, t = s/r ∈ Q.

It follows that

(x, y) = r2 − s2

r2 + s2,2rs. r2 + s2, s, r ∈ Z.

Here we take 0 < s < r, gcd(r, s) = 1. If in addition, r and s have opposite parities then the quotients will be in lowest terms, so that

(a, b, c) = (r2 − s2, 2rs, r2 + s2).

RATIONAL PARAMETRIZATION OF CONICS 1. The General Situation ...

PDFpeople.reed.edu › ~jerry › conics

Hello Adriana,

many thanks for your answer, it should help a lot.

Regards,

Gabriel