We can obviously see from the given conditions that both sequences $x_{1}^{2},x_{2}^{2},...$ and $\sqrt{y_{1}},\sqrt{y_{2}},...$ are strictly increasing arithmetic progressions. Write $x_{n}^{2}=x^{2}+(n-1)a$ and $\sqrt{y_{n}}=\sqrt{y}+(n-1)b,$ where $x=x_{1},\;\;y=y_{1},$ and $a,b>0.$ We can easily see now $$y_{n}=(\sqrt{y}+\frac{b(x_{n}^{2}-x^{2})}{a})^{2}=(Ax_{n}^{2}+B)^{2}$$where $A=\frac{b}{a},\;B=\sqrt{y}-\frac{bx^{2}}{a}.$ Now define $f(t)=\frac{(At^{2}+B)^{2}}{t}$ for $t>0,$ the collinearity condition would imply that $f(x)=f(x_{2016}).$ We claim that now $f(x_{n})<f(x)\;\forall \;2\le n\le 2015$ (in particular, all the points $A_{2},A_{3},...,A_{2015}$ lie to the right of $d$). Indeed, fix an $2\le n\le 2015,$ then $x_{1}<x_{n}<x_{2016}$, and so $$\frac{f(x_{n})-f(x)}{x_{n}-x}=A^{2}(x_{n}^{2}+x_{n}x+x^{2})+2AB-\frac{B^{2}}{x_{n}x}<A^{2}(x_{2016}^{2}+x_{2016}x+x^{2})+2AB-\frac{B^{2}}{x_{2016}x}=\frac{f(x_{2016})-f(x)}{x_{2016}-x}=0$$as claimed. So we are done.