JEE Main 2022 · 26 June · morning · Single correct · +4 / −1

Let the normal at the point on the parabola y² = 6x pass through the point (5, -8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :

  1. (a) -3
  2. (b) -9/4
  3. (c) -5/2
  4. (d) -2
Answer and solution

Answer (b) -9/4

Solution

Maths sir · 5 steps

Write P as (at², 2at) and make its normal pass through (5,-8). That cubic in t fixes P.

  1. Read off a. Compare y² = 6x with y² = 4ax: 4a = 6, so a = 3/2. The directrix is x = -a = -3/2, and a general point is P(at², 2at).

    Normal at t: y = -tx + 2at + at³

  2. Normal through point. Put (5, -8) and a = 3/2: -8 = -5t + 3t + 3/2t³. Multiply by 2 to get 3t³ - 4t + 16 = 0.
  3. Find P. t = -2 works: -24 + 8 + 16 = 0. The cubic factors as (t+2)(3t² - 6t + 8), and the quadratic has discriminant 36 - 96 < 0, so t = -2 is the only root. Then P = (6, -6).
  4. Tangent meets directrix. The tangent at t is ty = x + at², which here is -2y = x + 6. Put x = -3/2: -2y = 9/2.
  5. So. y = -9/4, so the ordinate of Q is -9/4, option (b).

The idea to keep. For a normal through a given point, use the parametric normal y = -tx + 2at + at³. Substituting the point gives a cubic in t, and each real root is one foot of a normal.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2022.