Question
If the normal to the parabola $y^2 = 4ax $ at the point $(at^2,2at)$ cuts the parabola again at $(aT^2,aT)$ , then prove that $T^2 \ge 8$.
Answer
Normal at $(at^2,2at)$ is $y+tx=2at+at^3$, which cuts the parabola again at $(aT^2,2aT)$, resulting in the following: $$ 2a(T-t)=-at(T^2-t^2)$$ $$ \implies 2=-t(T+t)$$ $$ \implies t^2+tT+2=0$$
My doubt starts here:
Since $t$ is real, $T^2-8 \ge 0$ which proves $T^2 \ge 8$.
But if $t$ is real does that necessarily imply that any equation it forms has only real roots?