(S) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}}\right)$
If a tangent to a suitable conic (Column 1) is found to be $y = x + 8$ and its point of contact is $(8, 16)$, then which of the following options is the only CORRECT combination? [A] (I) (ii) (Q) [B] (II) (iv) (R) [C] (III) (i) (P) [D] (III) (ii) (Q)
Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.
\begin{tabular}{|l|l|l|l|}
\hline
& Column 1 & Column 2 & Column 3 \\
\hline
& (I) $x^2 + y^2 = a^2$ & (i) $my = m^2x + a$ & (P) $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$ \\
\hline
(II) & $x^2 + a^2y^2 = a^2$ & (ii) $y = mx + a\sqrt{m^2+1}$ & (Q) $\left(\frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}}\right)$ \\
\hline
(III) & $y^2 = 4ax$ & (iii) $y = mx + \sqrt{a^2m^2 - 1}$ & (R) $\left(\frac{-a^2m}{\sqrt{a^2m^2+1}}, \frac{1}{\sqrt{a^2m^2+1}}\right)$ \\
\hline
(IV) & $x^2 - a^2y^2 = a^2$ & (iv) $y = mx + \sqrt{a^2m^2+1}$ & (S) $\left(\frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}}\right)$ \\
\hline
\end{tabular}
If a tangent to a suitable conic (Column 1) is found to be $y = x + 8$ and its point of contact is $(8, 16)$, then which of the following options is the only CORRECT combination?
[A] (I) (ii) (Q)
[B] (II) (iv) (R)
[C] (III) (i) (P)
[D] (III) (ii) (Q)