JEE Main 2020 · 6 September · morning · Single correct · +4 / −1

Let L₁ be a tangent to the parabola y² = 4(x + 1)
and L₂ be a tangent to the parabola y² = 8(x + 2)
such that L₁ and L₂ intersect at right angles. Then L₁ and L₂ meet on the straight line :

  1. (a) x + 3 = 0
  2. (b) x + 2y = 0
  3. (c) x + 2 = 0
  4. (d) 2x + 1 = 0
Answer and solution

Answer (a) x + 3 = 0

Solution

Maths sir · 4 steps

Write both tangents in slope form, make the slopes m and -1/m, and subtract to kill y.

  1. Slope form. For Y² = 4aX the tangent with slope m is Y = mX + a/m. The first parabola has a = 1 with X = x + 1, so L₁ is y = m(x+1) + 1/m.
  2. Perpendicular tangent. The second parabola has a = 2 with X = x + 2. Since L₂ ⊥ L₁, its slope is -1/m, so L₂ is y = -1/m(x+2) - 2m.
  3. Eliminate y. At the meeting point both right sides are equal. Bring everything to one side and group the terms.

    x(m + 1/m) + 3(m + 1/m) = 0

  4. So. Since m + 1/m ≠ 0, we get x = -3 for every m. The tangents always meet on x + 3 = 0, which is option (a).

The idea to keep. For a locus of meeting points of two tangents, write each in slope form with one parameter m and eliminate y. If m drops out completely, you have found the fixed line.

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

Study Parabola

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