Let the equation of the two tangents at the points $A(t_1)$ and $B(t_2)$ be $yt_1=x+at_1^2$ and $yt_2=x+at_2^2$
Subtract both the equations $y (t_1-t_2)=a(t_1^2-t_2)y(t_1 -t_2) =a(t_1-t_2)$
Hence $y=a(t_1+t_2)$ Substituting in equal $a(t_1+t_2) t_1=x+at_1^2$
Therefore at $1^2 +t_1 t_2 =x +t_1^2$
Therefore $x=t_1t_2$