2

By a linear transformation, a conic known by its implicit equation can be put in a reduced form such as that of a unit circle $x^2+y^2=1$. The latter can be represented by rational parametric equations,

$$\begin{cases}x=\dfrac{2t}{t^2+1},\\y=\dfrac{t^2-1}{t^2+1}.\end{cases}$$

Coming back to the original equation, we obtain two other rational expressions with quadratic numerator and denominator.

My question is: is there a direct way to turn a general conic to rational parametric equations of the second degree without resorting to centering and reduction of the conic ?

Update:

An easy solution is obtained by a simple change of variable such as $y=z+mx$. By substitution, the quadratic terms become

$$ax^2+2bxy+cy^2=ax^2+2bx(z+mx)+c(z+mx)^2$$ where the coefficient of $x^2$ is $a+2bm+cm^2$ and can be canceled (when there is a real solution). Then $x$ can be expressed as a rational expression in $z$, and so can $y$.

Easy, but ugly, and still requiring the roots of a quadratic equation :)

  • Some restrictions: Within affine geometry, if the linear transformation is required to be real, only ellipses can be reduced to circles. If the underlying field is $\mathbb{Q}$, the reduction to the unit circle can only be done if the ellipse contains a rational point, e.g. not for $x^2+y^2=3$. – ccorn Apr 06 '18 at 12:05
  • @ccorn: I said "such as" to be short and we are in the reals. –  Apr 06 '18 at 12:06
  • 1
    Diophantus's method: Given a point $P$ on the conic and its implicit equation, calculate the other point of intersection with a line of slope $t$ through $P$. If you do not have the point $P$ a priori, you need to calculate one, e.g. by intersecting with a suitable line, but that involves at least one square root. – ccorn Apr 06 '18 at 12:22
  • @ccorn: indeed, I have found several articles that seem to converge to that approach. I was hoping to find a simple relation between the implicit and parametric coefficients. –  Apr 06 '18 at 12:54
  • If $ax^2+2bxy+cy^2+2dx+2ey+f=0$ and $(x_1,y_1)$ is a rational point, $x={\frac {c{t}^{2}{x_{{1}}}^{2}+ \left( -2,y_{{1}}x_{{1}}c-2,ex_{{1}} \right) t+c{y_{{1}}}^{2}+f+2,ey_{{1}}}{ \left( c{t}^{2}+a+2,bt \right) x_{{1}}}}$

    $y=-{\frac { \left( y_{{1}}x_{{1}}c+2,ex_{{1}}+2,{x_{{1}}}^{2}b \right) {t}^{2}+ \left( -f-2,y_{{1}}x_{{1}}b+{x_{{1}}}^{2}a-2,ey_{{ 1}}-c{y_{{1}}}^{2} \right) t-y_{{1}}x_{{1}}a}{ \left( c{t}^{2}+a+2,bt \right) x_{{1}}}}$. Found using the method from this answer.

     – Jan-Magnus Økland Apr 07 '18 at 12:35
  • Thanks. $(x_1,y_1)$ doesn't have to be rational, I am not after rational numbers but rational expressions. The denominator is related to the characteristic polynomial of the quadratic subarray, which is not a surprise. Now finding the point seems as "difficult" as solving for the Eigenvalues. –  Apr 07 '18 at 12:47
  • Trying to understand. Like a conic $ \dfrac{x}{a} = \dfrac{2mn}{m^2+n^2},,\dfrac{y}{b} = \dfrac{m^2-n^2}{m^2+n^2} ? $Pythagorean triplets $(m.n)$ – Narasimham Nov 02 '18 at 06:22

0 Answers0