If P ( h, k ) be point on the parabola x = 4 y ^ 2 , which is...

Given parabola is

$x = 4{y}^{2}$

$⇒ {y}^{2} = \dfrac{1}{4}x$

$⇒ {y}^{2} = 4 \times \dfrac{1}{16} \times x$

Equation of normal is

$y = - tx + 2at + a{t}^{3}$

$⇒ y = - tx + \dfrac{2}{16}t + \dfrac{1}{16}{t}^{3}$

It passes through $Q\left(0, 33\right)$, so

$33 = \dfrac{t}{8} + \dfrac{{t}^{3}}{16}$

$⇒ {t}^{3} + 2t - 528 = 0$

$⇒ \left(t - 8\right)\left({t}^{2} + 8t + 66\right) = 0$

$⇒ t = 8$

So,

$P ≡ \left(h, k\right) ≡ \left(a{t}^{2}, 2at\right) ≡ ( 4, 1 )$

Now, we have parabola

${y}^{2} = 4\left(x + y\right)$

$⇒ {y}^{2} - 4y = 4x$

$⇒ {\left(y - 2\right)}^{2} = 4\left(x + 1\right)$

Equation of directrix is

$x + 1 = - 1$

$⇒ x = - 2$

So, distance of point \left(4, 1\right)$from the line$x + 2 = 0$is</p><p>$= \left|\dfrac{4 + 2}{1}\right| = 6$ units