Let T be the tangent to the parabola $y^{2} = 16x$ at the point $(64, 32)$. Let L be the tangent to the same parabola at another point $(x_{1}, y_{1})$ on the parabola. If L and T are perpendicular to each other, then the distance between the point $(x_{1}, y_{1})$ and the focus of the parabola, is (1) $\frac{15}{4}$ (2) $4$ (3) $\frac{17}{4}$ (4) $5$

Let T be the tangent to the parabola y² = 16x at the point (64, 32). Let L be the tangent to the same parabola at another point (x₁, y₁) on the parabola. If L and T are perpendicular to each other, then the distance between the point (x₁, y₁) and the focus of the parabola, is $\text{Parabola: } y^2 = 16x, \ \text{point } (64, 32)$ $T:\ \text{tangent at } (64, 32)$ $L:\ \text{tangent at } (x_1, y_1), \qquad L \perp T$ $\text{To find: distance from } (x_1, y_1) \text{ to the focus}$ $y^2 = 16x \ \Rightarrow\ 4a = 16 \Rightarrow a = 4$ $\text{Focus } S = (4, 0), \quad \text{directrix } x = -4$ $(64, 32) \text{ on } y^2 = 4ax:\ 2at = 32,\ at^2 = 64$ $8t = 32 \Rightarrow t = 4 \quad (\text{and } 4 \cdot 16 = 64\ \checkmark)$ $\text{Tangent at } t:\ ty = x + at^2, \quad \text{slope} = \dfrac{1}{t}$ $T:\ 4y = x + 64 \Rightarrow \text{slope}_T = \dfrac{1}{4}$ $\text{slope}_T \cdot \text{slope}_L = -1 \Rightarrow \dfrac{1}{4} \cdot \dfrac{1}{s} = -1$ $s = -\dfrac{1}{4}$ $x_1 = as^2 = 4 \cdot \dfrac{1}{16} = \dfrac{1}{4}$ $y_1 = 2as = 8 \cdot \left(-\dfrac{1}{4}\right) = -2$ $\Rightarrow (x_1, y_1) = \left(\dfrac{1}{4}, -2\right)$ $L:\ sy = x + as^2 \Rightarrow -\dfrac{1}{4}y = x + \dfrac{1}{4}$ $y = -4x - 1$ $T \cap L:\ 4y = x + 64,\quad y = -4x - 1$ $x = -4,\ y = 15 \Rightarrow \text{lies on directrix } x = -4$ $\text{Focal distance} = \text{distance to the directrix}$ $= x_1 - (-a) = x_1 + a$ $d = x_1 + a = \dfrac{1}{4} + 4 = \dfrac{17}{4}$ $d = \sqrt{\left(4 - \dfrac{1}{4}\right)^2 + (0 + 2)^2} = \sqrt{\dfrac{225}{16} + 4}$ $= \sqrt{\dfrac{289}{16}} = \dfrac{17}{4}$ $\boxed{d = \dfrac{17}{4}} \quad \text{option (3)}$