Remove the original intersection before using a discriminant

For a genuine parabola assume $a\ne0$; if $a=0$ the parametrisation collapses to a point and the stated tangent/normal problem is not defined. For $t\ne0$, $dy/dx=2a/(2at)=1/t$, so the normal slope is $-t$. Thus $y-2at=-t(x-at^2)$, or $y+tx=at^3+2at$. At $t=0$ the tangent is vertical and the normal is $y=0$, also given by this formula; it does not meet the parabola at a second point.

$2aq+t(aq^2)=at^3+2at$. Dividing by $a\ne0$ and rearranging gives $t^3+2t-q^2t-2q=0$.

Since the two distinct points lie on the normal, $(2at-2aq)/(at^2-aq^2)=-t$. After cross-multiplying and dividing by $a$, $2t-2q=-t^3+q^2t$, giving the same cubic. The finite normal cannot join two distinct points with the same x-coordinate, so the secant denominator is nonzero under these hypotheses.

For $F(t)=t^3+2t-q^2t-2q$, $F(q)=0$. Write $F(t)=(t-q)(t^2+bt+2)$; the $t^2$ coefficient gives $b=q$. Hence $F(t)=(t-q)(t^2+qt+2)$.

$t^3-q^2t+2t-2q=t(t^2-q^2)+2(t-q)=(t-q)[t(t+q)+2]=(t-q)(t^2+qt+2)$. A second distinct point has $q\ne t$, so $t^2+qt+2=0$.

Considering $t^2+qt+2=0$ as a quadratic in the real parameter $t$, its discriminant is $q^2-8\ge0$. Thus $q^2\ge8$. Equality is possible: $q=\pm2\sqrt2$, $t=-q/2$, which are distinct real parameters.