Distance Between Point on Parabola and Focus for Perpendicular Tangents

To find the distance between the point $(x_1, y_1)$ and the focus of the parabola, we proceed step-by-step:

Step 1: Identify the parameters of the parabola The given equation of the parabola is: $y^2 = 16x$

Comparing this with the standard equation of a parabola $y^2 = 4ax$, we get: $4a = 16 \implies a = 4$

The coordinates of the focus $S$ of the parabola are: $S = (a, 0) = (4, 0)$

Step 2: Find the slope of the tangent $T$ The parametric coordinates of any point on the parabola $y^2 = 4ax$ are given by $(at^2, 2at)$.

For the point $(64, 32)$ on the parabola: $2at = 32 \implies 2(4)t = 32 \implies 8t = 32 \implies t = 4$

The equation of the tangent at a parameter $t$ has a slope $m = \frac{1}{t}$. Therefore, the slope of tangent $T$, denoted as $m_T$, is: $m_T = \frac{1}{4}$

Step 3: Determine the point $(x_1, y_1)$ for the perpendicular tangent $L$ Let tangent $L$ touch the parabola at $(x_1, y_1)$ corresponding to parameter $t_1$. Its slope $m_L$ is: $m_L = \frac{1}{t_1}$

Since tangent $L$ is perpendicular to tangent $T$, the product of their slopes is $-1$: $m_L \cdot m_T = -1$ $\frac{1}{t_1} \cdot \frac{1}{4} = -1 \implies \frac{1}{t_1} = -4 \implies t_1 = -\frac{1}{4}$

Now, we calculate the coordinates $(x_1, y_1)$: $x_1 = a t_1^2 = 4 \cdot \left(-\frac{1}{4}\right)^2 = 4 \cdot \frac{1}{16} = \frac{1}{4}$ $y_1 = 2 a t_1 = 2 \cdot 4 \cdot \left(-\frac{1}{4}\right) = -2$

Thus, the point is $(x_1, y_1) = \left(\frac{1}{4}, -2\right)$.

Step 4: Calculate the distance from the point $(x_1, y_1)$ to the focus By the definition of a parabola, the distance of any point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ from its focus (focal distance) is given by: $\text{Distance} = x_1 + a$

Substituting $x_1 = \frac{1}{4}$ and $a = 4$: $\text{Distance} = \frac{1}{4} + 4 = \frac{17}{4}$

(Alternatively, using the distance formula between $\left(\frac{1}{4}, -2\right)$ and $(4, 0)$:) $\text{Distance} = \sqrt{\left(4 - \frac{1}{4}\right)^2 + (0 - (-2))^2} = \sqrt{\left(\frac{15}{4}\right)^2 + 2^2} = \sqrt{\frac{225}{16} + 4} = \sqrt{\frac{289}{16}} = \frac{17}{4}$

Thus, the required distance is $\frac{17}{4}$.

Correct Option: C