I am looking for a generalisation of the solution of $x,y$ wich are rational numbers,they could be infinite,how can i find such solutions,integer solutions are obvious
I have found that
$\Delta=-11,u=1,d=-11$
also
$(x+\frac{3+\sqrt{-11}y}{2})(x+\frac{3-\sqrt{-11}y}{2})=4$
On
This is a variation of Hecke's answer $$(2x+3y)^2+11y^2=4^2$$ $$\left(\frac{2x}y+3\right)^2+11=\left(\frac4y\right)^2$$ $$11=\left(\frac4y\right)^2-\left(\frac{2x}y+3\right)^2$$ $$11=\left(\frac{4-2x}y-3\right)\left(\frac{4+2x}y+3\right)$$ $$\frac{4-2x}y-3=t, \frac{4+2x}y+3=\frac{11}t$$ $$\implies x = -\frac{2(t^2+6t-11)}{t^2+11}, y = \frac{8t}{t^2+11}, t\neq0$$
My comment contained an imprecision. This works for every equation that can be transformed into the form: $$(ax+by)^2+cy^2=\left(\frac pq\right)^2$$ Where $a,b,c,p,q\in\mathbb Z, q\neq 0$
We have the simple solution $x=2$, $y=0$. Now take the line though $(2,0)$ with slope $m$, and find where it meets the ellipse. That will give a rational parametrization of the ellipse.
Details: The line has equation $y=m(x-2)$. Substitute. We get $x^2+3xm(x-2)+5m^2(x-2)^2=16$. This simplifies to $(1+3m+5m^2)x^2-(6m+20m^2)x+20m^2-4=0$. The product of the roots is $\frac{20m^2-4}{1+3m+5m^2}$. But one of the roots is $2$, so the other is $\frac{10m^2-2}{1+3m+5m^2}$. Now we can compute the corresponding $y$. Any rational $m$ will give us a rational solution of the original equation.
Remark: The procedure was general. In particular, let $ax^2+bxy+cy^2=d$ be an ellipse with $a,b,c,d$ rational. If we know a rational point on the ellipse, then we can find a parametric expression for all rational points. The most important special case is the circle.
And the basic idea generalizes, importantly, to elliptic curves.