Question

P and Q are two distinct points on the parabola, $y^2=4x$, with parameters t and $t_1$ respectively. If the normal at P passes through Q, then the minimum value of $t_1^2$ is :

Answer 1

$P(t^2,2t)Q(t_1^2,2t_1)$

Normal in parametric form at $P$:

$xt-y=2t+t^3\to$ passes through $Q$

$t{t}_1-2t_1=2t+t_3$

$\Rightarrow 2(t+t_1)=t(t_1^2-t^2)$

$\Rightarrow (t+t_1)=0$ OR $t(t_1-t)=2$

$\Rightarrow t_1=t+\frac{2}{t}\ge 2\sqrt{2}$ $[\because \frac{t+\frac{2}{t}}{2}\ge \sqrt{2}]$

$(t_1^2{)}_{min}=(2\sqrt{2}{)}^2=8$