Matching Tangent Equation and Contact Point for Parabola

To determine the correct combination, we analyze the given information:

1. Equation of the tangent: $y = x + 8$ Comparing this with the slope-intercept form $y = mx + c$, we get the slope $m = 1$ and the $y$-intercept $c = 8$.

2. Point of contact: $(x_1, y_1) = (8, 16)$

Let's test the conic given in Column 1 (III), which is the parabola $y^2 = 4ax$:

• Finding the parameter $a$: Since the point of contact $(8, 16)$ lies on the parabola $y^2 = 4ax$, we substitute $x = 8$ and $y = 16$: $16^2 = 4a(8) \implies 256 = 32a \implies a = 8$

• Checking Column 2 (i): The equation of the tangent to the parabola $y^2 = 4ax$ in terms of slope $m$ is: $y = mx + \frac{a}{m} \implies my = m^2 x + a$ This directly corresponds to entry (i) in Column 2. Substituting $a = 8$ and $m = 1$: $1 \cdot y = 1^2 \cdot x + 8 \implies y = x + 8$ This matches the given tangent equation.

• Checking Column 3 (P): The point of contact for the tangent with slope $m$ on the parabola $y^2 = 4ax$ is given by: $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$ This directly corresponds to entry (P) in Column 3. Substituting $a = 8$ and $m = 1$: $\left(\frac{8}{1^2}, \frac{2 \times 8}{1}\right) = (8, 16)$ This matches the given point of contact.

Thus, the only correct combination is (III) (i) (P).

Correct Answer: C