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Let a tangent be drawn to the ellipse $${{{x^2}} \over {27}} + {y^2} = 1$$ at $$(3\sqrt 3 \cos \theta ,\sin \theta )$$ where $$0 \in \left( {0,{\pi \over 2}} \right)$$. Then the value of $$\theta$$ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

JEE · Math · previous-year question

  1. A.$${{\pi \over 6}}$$correct
  2. B.$${{\pi \over 3}}$$
  3. C.$${{\pi \over 8}}$$
  4. D.$${{\pi \over 4}}$$

Answer

A. $${{\pi \over 6}}$$

Explanation

Tangent = $${x \over {3\sqrt 3 }}\cos \theta + y\sin \theta = 1$$ x-intercept = $${3\sqrt 3 }$$ sec$$\theta$$ y-intercept = cosec$$\theta$$ sum = $${3\sqrt 3 }$$ sec$$\theta$$ + cosec$$\theta$$ = f($$\theta$$) $$\theta$$$$\in$$$$\left( {0,{\pi \over 2}} \right)$$ $$ \Rightarrow $$ f'($$\theta$$) = $${3\sqrt 3 }$$ sec$$\theta$$tan$$\theta$$ $$-$$ cosec$$\theta$$ cot$$\theta$$ = 0 $$ \Rightarrow $$ $${{3\sqrt 3 \sin \theta } \over {{{\cos }^2}\theta }} = {{\cos \theta } \over {\sin \theta }}$$ $$ \Rightarrow {\tan ^3}\theta = {\left( {{1 \over {\sqrt 3 }}} \right)^3}$$ $$ \Rightarrow \tan \theta = {1 \over {\sqrt 3 }}$$ $$ \Rightarrow $$ $$\theta$$ = $${{\pi \over 6}}$$ also f'($$\theta$$) changes sign $$-$$ to + hence minimum.

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