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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Parabola

If the chord joining the points P_ 1 (x_ 1 , y_ 1 ) and P_ 2 (x_ 2...

If the chord joining the points P1(x1,y1)\mathrm{P}_{1}\left(x_{1}, y_{1}\right) and P2(x2,y2)\mathrm{P}_{2}\left(x_{2}, y_{2}\right) on the parabola y2=12xy^{2}=12 x subtends a right angle at the vertex of the parabola, then x1x2−y1y2x_{1} x_{2}-y_{1} y_{2} is equal to

Options

  1. A284
  2. B280
  3. C288
  4. D292

Correct answer

C. 288

Step-by-step solution

Parabola y2=12xy^2 = 12x has 4a=124a = 12, so a=3a = 3.
Parametric points: P1=(3t12,6t1)P_1 = (3t_1^2, 6t_1), P2=(3t22,6t2)P_2 = (3t_2^2, 6t_2).
Chord subtends right angle at vertex O(0,0)O(0,0): OP1⃗⋅OP2⃗=0\vec{OP_1} \cdot \vec{OP_2} = 0.
9t12t22+36t1t2=09t_1^2 t_2^2 + 36t_1 t_2 = 0
⇒t1t2(t1t2+4)=0\Rightarrow t_1 t_2(t_1 t_2 + 4) = 0
⇒t1t2=−4\Rightarrow t_1 t_2 = -4.
x1x2−y1y2=9(t1t2)2−36(t1t2)x_1 x_2 - y_1 y_2 = 9(t_1 t_2)^2 - 36(t_1 t_2)
=9(16)−36(−4)=144+144=288= 9(16) - 36(-4) = 144 + 144 = 288.
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