JEE Advanced 2026 — Paper 2 — MATH-I solution

JEE Advanced2026Paper 2MATH-I

Let T be the tangent to the parabola $y^{2} = 16 x$ 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

A
$\frac{15}{4}$
B
$4$
C
$\frac{17}{4}$
D
$5$

Solution

For $y^{2} = 16 x$ , the tangent at $(64, 32)$ uses $T : y y_{0} = 8 (x + x_{0})$ , giving $32 y = 8 (x + 64)$ , that is $x - 4 y + 64 = 0$ with slope $\frac{1}{4}$ .

Parametric points on $y^{2} = 16 x$ may be written as $(4 t^{2}, 8 t)$ with tangent slope $\frac{1}{t}$ .

For L $⊥$ T, the product of slopes is $- 1$ , so $\frac{1}{t} \cdot \frac{1}{4} = - 1$ , giving $t = - \frac{1}{4}$ .

Therefore $(x_{1}, y_{1}) = (\frac{1}{4}, - 2)$ , and the focus is $(4, 0)$ .

Distance $= (4 - \frac{1}{4})^{2} + 4 = \frac{225}{16} + 4 = \frac{289}{16} = \frac{17}{4}$ .

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