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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    If the tangent at a point on the ellipse\[\frac{{{x}^{2}}}{27}+\frac{{{y}^{2}}}{3}=1\]meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is :   JEE Main Online Paper (Held On 09 April 2016)

    A) 9             

    B) \[\frac{9}{2}\]

    C) \[9\sqrt{3}\]                                      

    D) \[3\sqrt{3}\]

    Correct Answer: A

    Solution :

                    Let \[P(3\sqrt{3}\cos \theta ,\sqrt{3}\sin \theta )\] \[\therefore \]tangent is\[\frac{x}{3\sqrt{3}}\cos \theta +\frac{y}{\sqrt{3}}\sin \theta =1\] \[\Rightarrow \]\[A(3\sqrt{3}\sec \theta ,0)\]    \[B(0,\sqrt{3}\sec ec\theta )\] \[\therefore \]Area of \[\Delta OAB=\frac{1}{2}OA.OB\] \[=\frac{1}{3}(3\sqrt{3}\sec \theta .\sqrt{3}\cos ec\theta )\] \[=\frac{9}{2\sin \theta \cos \theta }=\frac{9}{\sin 2\theta }\] \[\therefore \]minimum area of \[\Delta OAB=\frac{9}{1}=9\]                


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