JEE Main 2022 · 26 July · evening · Single correct · +4 / −1

The acute angle between the pair of tangents drawn to the ellipse 2 x²+3 y²=5 from the point (1,3) is :

  1. (a) tan ⁻¹(16/(7 √5))
  2. (b) tan ⁻¹(24/(7 √5))
  3. (c) tan ⁻¹(32/(7 √5))
  4. (d) tan ⁻¹((3+8 √5)/35)
Answer and solution

Answer (b) tan ⁻¹(24/(7 √5))

Solution

Maths sir · 5 steps

Use the tangency condition to get one quadratic in the slope; its two roots are the two tangents.

  1. Standard form. Divide by 5 to get x²/(5/2) + y²/(5/3) = 1, so a² = 5/2 and b² = 5/3. The line y = mx + c touches this ellipse when c² = a²m² + b².
  2. Through the point. A tangent through (1, 3) has c = 3 - m. Put this into the condition and multiply through by 6.

    (3-m)² = 5/2m² + 5/3 ⇒ 9m² + 36m - 44 = 0

  3. Sum and product. The roots m₁, m₂ are the slopes of the two tangents, with m₁ + m₂ = -4 and m₁ m₂ = -44/9. Then (m₁ - m₂)² = 16 + 176/9 = 320/9, so |m₁ - m₂| = 8√5/3.
  4. Angle formula. Here 1 + m₁ m₂ = -35/9. For the acute angle take tan θ = |(m₁ - m₂)/(1 + m₁ m₂)| = (8√5/3)/(35/9) = 24√5/35 = 24/7√5.
  5. So. The acute angle is tan⁻¹(24/7√5), which is option (b).

The idea to keep. For tangents from an external point, put the point into c² = a²m² + b². You get a quadratic in m, and its sum and product of roots give the angle without finding either slope.

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

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Asked about 100 times in JEE Main since 2020. This question is from 2022.