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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Tangent and Normal

  • question_answer
    The normal to the curve \[x=a\text{ }(1+\cos \theta ),\,y=a\sin \theta \]at \['\theta '\] always passes through the fixed point                                        [AIEEE 2004]

    A)            (a, a)

    B)            (0, a)

    C)            (0, 0)

    D)            (a, 0)

    Correct Answer: D

    Solution :

               Slope of normal \[=\frac{-dx}{dy}=\frac{\frac{-d\left\{ a(1+\cos \theta ) \right\}}{d\theta }}{\frac{d(a\sin \theta )}{d\theta }}=\frac{a\sin \theta }{a\cos \theta }=\tan \theta \]            Now, the equation of normal at \[\theta \]is,                     \[y-a\sin \theta =\tan \theta [x-a(1+\cos \theta )]\]                    Clearly, this line passes through (a, 0).


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