P and Q are two distinct points on the parabola, y ^ 2 = 4x , with...

Given $$P\left({t}^{2}, 2t\right)$ and $Q\left({t}_{1}^{2}, 2{t}_{1}\right)$, are parametric points on parabola
${y}^{2} = 4x$
Slope of normal at $P = - t$

Since normal passes through $Q$,

$\to$ slope of normal at $P$ = slope of $PQ$

$\to - t = \dfrac{2\left({t}_{1} - t\right)}{{t}_{1}^{2} - {t}^{2}}$

$\to - t = \dfrac{2}{{t}_{1} + t}$

$\to {t}_{1} = - t - \dfrac{2}{t}$

$\to {t}_{1}^{2} = {t}^{2} + \dfrac{4}{{t}^{2}} + 4 . . . \left(1\right)$

{t}^{2} + \dfrac{4}{{t}^{2}} ≥ 2\sqrt{{t}^{2} . \dfrac{4}{{t}^{2}}} ≥ 4,

Using AM ≥ GM

∴ {t}_{1}^{2} ≥ 4 + 4\left(from\left(1\right)\right)

\to {t}_{1}^{2} ≥ 8

∴ Minimum value$= 8