JEE Main 2020 · 2 September · morning · Single correct · +4 / −1

A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola
x²/4 - y²/2 = 1 at the point ( x₁,y₁ ). Then x₁² + 5y₁² is equal to :

  1. (a) 5
  2. (b) 6
  3. (c) 10
  4. (d) 8
Answer and solution

Answer (b) 6

Solution

Maths sir · 4 steps

Write the tangent at (x₁, y₁), set its slope equal to 2, and the point drops out at once.

  1. Tangent at the point. For x²/4 - y²/2 = 1, the tangent at (x₁, y₁) is (x x₁)/4 - (y y₁)/2 = 1. Its slope is (x₁/4)/(y₁/2) = x₁/2y₁.
  2. Match the slope. The line 2x - y = 0 has slope 2, so x₁/2y₁ = 2, which gives x₁ = 4y₁.
  3. Point lies on curve. Put x₁ = 4y₁ into the hyperbola: 16y₁²/4 - y₁²/2 = 1, so 7y₁²/2 = 1. Hence y₁² = 2/7 and x₁² = 16y₁² = 32/7.

    x₁² + 5y₁² = 32/7 + 10/7 = 42/7 = 6

  4. So. x₁² + 5y₁² = 6, which is option (b).

The idea to keep. For a tangent with a given slope, write the tangent at (x₁, y₁), equate slopes to link x₁ and y₁, then use the fact that the point lies on the curve.

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

Study Hyperbola

Asked about 60 times in JEE Main since 2020. This question is from 2020.