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AIEEE 2003

JEE Main 2003 Mathematics question on Conic Sections

Tangent is drawn to ellipse x227+y2=1\frac{x^2}{27} + y^2 = 1 at (33cos⁡θ,sin⁡θ)(3\sqrt{3}\cos\theta, \sin\theta) (where, θ∈(0,π/2)\theta \in (0, \pi/2)). Then, the value of θ\theta such that the sum of intercepts on axes made by this tangent is minimum, is
  1. A.
    π3\frac{\pi}{3}
  2. B.
    π6\frac{\pi}{6}
  3. C.
    π8\frac{\pi}{8}
  4. D.
    π4\frac{\pi}{4}
Show answer and solution

Correct option: B

Solution

The tangent to x²/27 + y² = 1 at (3√3 cosθ, sinθ) is:

x·(3√3 cosθ)/27 + y·sinθ = 1, i.e., x cosθ/(3√3) + y sinθ = 1.

X-intercept = 3√3/cosθ, Y-intercept = 1/sinθ.

Sum S = 3√3/cosθ + 1/sinθ.

dS/dθ = 3√3 sinθ/cos²θ − cosθ/sin²θ = 0

⟹ 3√3 sin³θ = cos³θ ⟹ tan³θ = 1/(3√3) = 1/3³/² ⟹ tanθ = 1/√3 ⟹ θ = π/6.

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