Find a normal and distinguish its second curve intersection

$dx/dt=[4(1-t^2)+8t^2]/(1-t^2)^2=(4+4t^2)/(1-t^2)^2$, while $dy/dt=2t/(1-t^2)^2$. Thus $dy/dx=t/[2(1+t^2)]$, with the original exclusions $t\ne\pm1$.

$4/3=1/(1-t^2)$ gives $t=\pm1/2$. At $t=1/2$, x is $8/3$; at $t=-1/2$, x is $-8/3$. Hence P uses $t=1/2$. Its gradient is $1/5$, so the normal gradient is $-5$. Thus $y-4/3=-5(x-8/3)$, giving $3y=-15x+44$.

At parameter q, $3/(1-q^2)=-60q/(1-q^2)+44$. Since $q\ne\pm1$, multiply to obtain $44q^2+60q-41=0=(2q-1)(22q+41)$. The roots are $1/2$ (the existing P, rejected for “again”) and $-41/22$. Therefore t at Q is $-41/22$.