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JEE Main 202118 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2021 · 18 March · Evening Shift · Mathematics · Ellipse

Let a tangent be drawn to the ellipse x ^ 2 27 + y ^ 2 = 1 at ( 3 3...

Let a tangent be drawn to the ellipse x227+y2=1\dfrac{{x}^{2}}{27} + {y}^{2} = 1 at (33cosθ,sinθ)( 3\sqrt{3}cos\theta, sin\theta ) where θ∈(0,π2)\theta \in \left(0, \dfrac{\pi}{2}\right). Then the value of θ\theta such that the sum of intercepts on axes made by this tangent is minimum is equal to :Let a tangent be drawn to the ellipse x ^ 2 27 + y ^ 2 = 1 at ( 3 3 cosθ, sinθ ) where θ ∈ (0, (π)/(2)) . Then the

Options

  1. Aπ8\dfrac{\pi}{8}
  2. Bπ4\dfrac{\pi}{4}
  3. Cπ6\dfrac{\pi}{6}
  4. Dπ3\dfrac{\pi}{3}

Correct answer

C. π6\dfrac{\pi}{6}

Step-by-step solution

Equation of tangent be

xcosθ33+y⋅sinθ1=1,θ∈(0,π2)\dfrac{xcos\theta}{3\sqrt{3}} + \dfrac{y \cdot sin\theta}{1} = 1, \theta \in \left(0, \dfrac{\pi}{2}\right)

intercept on xx-axis

OA=33secθOA = 3\sqrt{3}sec\theta

intercept on yy-axis

OB=cosecθOB = cosec \theta

Now, sum of intercept

=33secθ+cosecθ=f(θ)= 3\sqrt{3}sec\theta + cosec \theta = f ( \theta ) let

f′(θ)=33secθtanθ−cosecθcotθf ' ( \theta ) = 3\sqrt{3}sec\theta tan\theta - cosec \theta cot\theta

=33sinθcos⁡2θ−cosθsin⁡2θ= 3\sqrt{3}\dfrac{sin\theta}{\cos^{2}\theta} - \dfrac{cos\theta}{\sin^{2}\theta}

=cosθsin⁡2θ⋅33⏟⊕[tan⁡3θ−133]=0⇒θ=π6= \mathop{\mathop{\dfrac{cos\theta}{\sin^{2}\theta} \cdot 3\sqrt{3}}_{⏟}}_{⊕}\left[\tan^{3}\theta - \dfrac{1}{3\sqrt{3}}\right] = 0 ⇒ \theta = \dfrac{\pi}{6}

Let a tangent be drawn to the ellipse x ^ 2 27 + y ^ 2 = 1 at ( 3 3 cosθ, sinθ ) where θ ∈ (0, (π)/(2)) . Then the

⇒⇒ at \theta = \dfrac{\pi}{6}, f\left(\theta\right) is minimum.

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