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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

  • question_answer
    The tangent to the curve \[x=a(\theta -\sin \theta ),\]\[y=a(1+\cos \theta )\] at the points \[\theta =(2k+1)\pi ,k\in I\] are parallel to

    A) \[y=x\]            

    B) \[y=-x\]

    C) \[y=0\]            

    D) \[x=0\]

    Correct Answer: C

    Solution :

    Given that, \[x=a(\theta -\sin \theta ),y=a(1+\cos \theta )\]On differentiating w.r.t. \[\theta \] respectively, we get \[\frac{dx}{d\theta }=a(1-\cos \theta ),\frac{dy}{d\theta }=a(-\sin \theta )\] \[\therefore \]\[\frac{dy}{dx}=\frac{-\sin \theta }{1-\cos \theta }\] For parallel curve, \[\frac{dy}{dx}=0\Rightarrow \theta =(2k+1)\pi .\] So, the tangent is parallel to x-axis ie, y = 0.


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