Step 1: Analyze and Match the Columns

First, we need to correctly match the conic in Column 1 with its corresponding tangent equation in Column 2 and point of contact in Column 3. We will assume standard forms for the conics, correcting for apparent typos in the question.

• Conic (I): Circle $x^2 + y^2 = a^2$ (assuming the typo $a$ should be $a^2$).

  • The standard equation of a tangent with slope $m$ to this circle is $y = mx \pm a\sqrt{m^2 + 1}$. Column 2, option (ii) is $y = mx + a\sqrt{m^2 + 1}$.
  • The point of contact for this tangent is $\left( \frac{-ma^2}{\pm a\sqrt{m^2+1}}, \frac{a^2}{\pm a\sqrt{m^2+1}} \right) = \left( \mp \frac{ma}{\sqrt{m^2+1}}, \pm \frac{a}{\sqrt{m^2+1}} \right)$. For the tangent with the '+' sign, the point is $\left( \frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}} \right)$. This matches Column 3, option (Q).
  • So, the correct combination for the circle is (I) $\rightarrow$ (ii) $\rightarrow$ (Q).

• Conic (III): Parabola $y^2 = 4ax$.

  • The standard equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$. This can be rewritten as $my = m^2x + a$, which matches Column 2, option (i).
  • The point of contact for this tangent is $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$. This matches Column 3, option (P).
  • So, the correct combination for the parabola is (III) $\rightarrow$ (i) $\rightarrow$ (P).

• Conic (II): Ellipse $\frac{x^2}{a^2} + y^2 = 1$ (assuming from $x^2 + a^2y^2 = a^2$).

  • The standard equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + 1}$. Column 2, option (iv) is $y = mx + \sqrt{a^2m^2 + 1}$.
  • The point of contact for this tangent is $\left( \frac{-a^2m}{\pm\sqrt{a^2m^2+1}}, \frac{1}{\pm\sqrt{a^2m^2+1}} \right)$. For the tangent with the '+' sign, the point is $\left( \frac{-a^2m}{\sqrt{a^2m^2+1}}, \frac{1}{\sqrt{a^2m^2+1}} \right)$. This matches Column 3, option (R).
  • So, the correct combination for the ellipse is (II) $\rightarrow$ (iv) $\rightarrow$ (R).

• Conic (IV): Hyperbola $\frac{x^2}{a^2} - y^2 = 1$ (assuming from $x^2 - a^2y^2 = a^2$).

  • The standard equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - 1}$. Column 2, option (iii) is $y = mx + \sqrt{a^2m^2 - 1}$.
  • The point of contact is $\left( \frac{-a^2m}{\pm\sqrt{a^2m^2-1}}, \frac{-1}{\pm\sqrt{a^2m^2-1}} \right)$. For the tangent with the '+' sign, the point is $\left( \frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}} \right)$. This matches Column 3, option (S).
  • So, the correct combination for the hyperbola is (IV) $\rightarrow$ (iii) $\rightarrow$ (S).

Step 2: Use the Given Specific Information

We are given that a tangent to one of these conics is the line $y = x + 8$ and its point of contact is $(8, 16)$.

First, let's verify that the point $(8, 16)$ lies on the line $y = x + 8$. Substituting $x=8$ gives $y = 8 + 8 = 16$. The point lies on the line, so the information is consistent.

From the tangent equation $y = x + 8$, we can identify the slope $m=1$ and the y-intercept $c=8$.

Step 3: Test each Conic Combination

We will now check which of the matched combinations from Step 1 fits this specific data.

1. Parabola (III) $\rightarrow$ (i) $\rightarrow$ (P):

   • Conic: $y^2 = 4ax$.
   • Tangent: $y = mx + \frac{a}{m}$.
   • Point of contact: $\left( \frac{a}{m^2}, \frac{2a}{m} \right)$.
   • From the given tangent, $m=1$. The equation becomes $y = x + a$. Comparing this with $y = x + 8$, we get $a=8$.
   • Now, let's find the point of contact using $a=8$ and $m=1$: $\left( \frac{8}{1^2}, \frac{2(8)}{1} \right) = (8, 16)$This matches the given point of contact.
   • The conic is $y^2 = 4(8)x = 32x$. Let's check if the point $(8, 16)$ lies on it: $16^2 = 256$ and $32 \times 8 = 256$. The point is on the conic.
   • This combination is a perfect match for the given information.

2. Circle (I) $\rightarrow$ (ii) $\rightarrow$ (Q):

   • Tangent intercept $c = a\sqrt{m^2+1}$. With $m=1$ and $c=8$, we have $8 = a\sqrt{1^2+1} = a\sqrt{2}$, so $a = 4\sqrt{2}$.
   • Point of contact: $\left( \frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}} \right) = \left( \frac{-1(4\sqrt{2})}{\sqrt{2}}, \frac{4\sqrt{2}}{\sqrt{2}} \right) = (-4, 4)$.
   • This does not match $(8, 16)$.

3. Ellipse (II) $\rightarrow$ (iv) $\rightarrow$ (R):

   • Tangent intercept $c = \sqrt{a^2m^2+1}$. With $m=1$ and $c=8$, we have $8 = \sqrt{a^2(1)^2+1}$, so $64 = a^2+1$, which gives $a^2=63$.
   • Point of contact: $\left( \frac{-a^2m}{\sqrt{a^2m^2+1}}, \frac{1}{\sqrt{a^2m^2+1}} \right) = \left( \frac{-63(1)}{8}, \frac{1}{8} \right)$.
   • This does not match $(8, 16)$.

4. Hyperbola (IV) $\rightarrow$ (iii) $\rightarrow$ (S):

   • Tangent intercept $c = \sqrt{a^2m^2-1}$. With $m=1$ and $c=8$, we have $8 = \sqrt{a^2(1)^2-1}$, so $64 = a^2-1$, which gives $a^2=65$.
   • Point of contact: $\left( \frac{-a^2m}{\sqrt{a^2m^2-1}}, \frac{-1}{\sqrt{a^2m^2-1}} \right) = \left( \frac{-65(1)}{8}, \frac{-1}{8} \right)$.
   • This does not match $(8, 16)$.

Step 4: Conclusion

The only conic that fits the given tangent line and point of contact is the parabola, corresponding to the combination (III) $\rightarrow$ (i) $\rightarrow$ (P). Therefore, the only correct combination among the options is (III) (i) (P).

Looking at the options: A: (III) (i) (P) B: (I) (ii) (Q) C: (II) (iv) (R) D: (III) (ii) (Q)

Option A is the correct choice.