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Code $(D) : (P) - 2, (Q) - 4, (R) - 3, (S) - 1$
$(P)$ Equation of normal is $y = - t x + 2 a t + a t^{3} a t P (t)$
It intersect the curve again at point $Q (t_{1})$ on the parabola such that
$t_{2} = - t - \frac{2}{t}$
Again slope of $O P$ is $\frac{2}{t} = M_{O P}$
Also, slope of $O Q$ is $\frac{2}{t_{1}} = M_{O Q}$
Since $M_{O P} \cdot M_{O Q} = - 1 = \frac{4}{t t_{1}}$
$\Rightarrow t t_{1} = - 4$
$t (\begin{matrix} - t - \frac{2}{t} \end{matrix}) = - 4$
$\Rightarrow t^{2} = 2$
$(Q) P (1, 2), Q (4, 4), R (16, 8)$
Now, $a r (Δ P Q R) = 6 s q . u n i t s$
$(R)$ Equation of normal from any point $P (a m^{2}, - 2 m)$ is
$y = m x - 2 a m - a m^{3}$
It passes through $(\begin{matrix} \frac{11}{4}, \frac{1}{4} \end{matrix})$
$\Rightarrow 4 m^{3} + 8 m - 11 m + 1 = 0$
$\Rightarrow 4 m^{3} - 3 m + 1 = 0$
Now, $f (m) = 4 m^{3} - 3 m$
$\Rightarrow f^{'} (m) = 12 m^{2} - 3 = 0$
$\Rightarrow m \pm \frac{1}{2}$
Since $f (\begin{matrix} \frac{1}{2} \end{matrix}) f (\begin{matrix} - \frac{1}{2} \end{matrix}) < 0$ has $3$ normals are possible.
$(S)$ Since, normal at $P (t_{1})$ it meets the curve again at $(t_{2})$ , then
$t_{2} = - t_{1} - \frac{2}{t_{1}}$
Such that here normal at $P (1)$ meets the curve again at $Q (t)$
$\Rightarrow t = - 1 - \frac{1}{2} = - 3$

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