JEE Main 2020 · 8 January · evening · Single correct · +4 / −1

The length of the perpendicular from the origin, on the normal to the curve,
x² + 2xy – 3y² = 0 at the point (2,2) is

  1. (a) √2
  2. (b) 4√2
  3. (c) 2
  4. (d) 2√2
Answer and solution

Answer (d) 2√2

Solution

Maths sir · 3 steps

Factor the curve first: it is a pair of lines, so the tangent at (2,2) is the line it lies on.

  1. Factor the curve. x² + 2xy - 3y² = (x - y)(x + 3y), so the curve is the pair of lines y = x and x + 3y = 0. The point (2, 2) lies on y = x.
  2. Tangent and normal. For a straight line, the tangent at any point is the line itself, so the tangent at (2, 2) is y = x with slope 1. The normal has slope -1 and passes through (2, 2): y - 2 = -(x - 2), that is x + y - 4 = 0.

    d = (|0 + 0 - 4|)/√(1² + 1²) = 4/√2 = 2√2

  3. So. The perpendicular from the origin to the normal has length 2√2, which is option (d).

The idea to keep. Before differentiating a second-degree curve, check whether it factors into lines. For any line ax + by + c = 0, the distance from the origin is |c|/√(a² + b²).

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

Study Straight Lines

Asked about 130 times in JEE Main since 2020. This question is from 2020.