Point at infinity on quartic elliptic curve Elliptic curve defined by
$$E_1: y^2=7 x^4+x^3+x^2+x+3, P_1=(-1,3)$$
can be transformed to
$$E_2: v^2=u^3-\frac{250 u}{3}-\frac{1249}{27}$$
Substitutions used are:
$$\left(x\to \frac{15 u-9 v+217}{39 u+9 v+209},y\to \frac{9 \left(54 u^3+639 u^2-27 v^2+592 v-16501\right)}{(39 u+9 v+209)^2}\right)$$
$$\left(u\to \frac{2 \left(20 x^2+x+9 y+8\right)}{3 (x+1)^2},v\to -\frac{81 x^3+3 x^2+26 x y-3 x-10 y-33}{(x+1)^3}\right)$$
Then we can check, as an example, that point $P_2=(\frac{6}{7},-3)$ on $E_1$ correcponds with point $Q_2=(-\frac{2}{3},3)$ on $E_2$.
Questions:

  
*
  
*Point $P_{\infty}=(0,1,0)$ on $E_1$ corresponds with what point on $E_2$?
  
*Point $Q_{\infty}=(0,1,0)$ on $E_2$ corresponds with what point on $E_1$?
  
*Point $P_1=(-1,3)$ on $E_1$ corresponds with what point on $E_2$?
  
*Point $Q_1=(-\frac{71}{9},\frac{296}{27})$ on $E_2$ corresponds with what point on $E_1$?
  

EDIT:
Maybe it was not clear, but for points $P_2=(\frac{6}{7},-3)$ and $Q_2=(-\frac{2}{3},3)$ I used the substitutions to verify they correspond to each other. For points in my question the same method did not work for me because of singularities (division by zero).


$$E_1(-1,3)\to E_2(0,1,0)$$
$$E_1(-1,-3)\to E_2(-\frac{71}{9},-\frac{296}{27})$$
$$E_1(-\frac{5413}{16069},-\frac{434267883}{258212761})\to E_2(-\frac{71}{9},\frac{296}{27})$$
$$E_1(1,-\sqrt{7},0)\to E_2(\frac{40}{3}-6 \sqrt{7},-81+26 \sqrt{7})$$
$$E_1(1,\sqrt{7},0)\to E_2(\frac{40}{3}+6 \sqrt{7},-81-26 \sqrt{7})$$
