JEE Advanced
Mathematics
Parabola
2026
JEE Advanced 2026 (Paper 2)
JEE Advanced Mathematics Question (2026) — Solution
Question
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
Options
- A. 15 4
- B. 4
- C. 17 4
- D. 5
Step-by-step solution
The equation of the parabola is y^2 = 16x. Comparing with y^2 = 4ax, we get a = 4. Let the parametric coordinates of the point (64, 32) be (at_1^2, 2at_1). 2at_1 = 32 8t_1 = 32 t_1 = 4 The slope of the tangent T at t_1 is m_1 = 1 t_1 = 1 4 . Let the tangent L be at the point (x_1, y_1) with parameter t_2. Its slope is m_2 = 1 t_2 . Since T and L are perpendicular, m_1 m_2 = -1. 1 4 1 t_2 = -1 t_2 = - 1 4 The x-coordinate of the point (x_1, y_1) is x_1 = at_2^2 = 4 (- 1 4 )^2 = 1 4 . The distance of a point (x_1, y_1) on the parabola from the focus is given by its focal distance x_1 + a. Distance = 1 4 + 4 = 17 4 Answer: 17 4
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