JEE Main & Advanced AIEEE Solved Paper-2004

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

    A) (a, 0)  

    B)                        (0, a)     

    C)        (0, 0)     

    D)        (a, a)

    Correct Answer: A

    Solution :

    Equation of normal at\[({{x}_{1}},{{y}_{1}})\]is \[(y-{{y}_{1}})=-\frac{1}{{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},{{y}_{1}})}}}(x-{{x}_{1}})\] Given,      \[x=a(1+\cos \theta ),y=a\sin \theta \] On differentiating w.r.t.\[\theta \]respectively, we get \[\frac{dx}{d\theta }=a(-\sin \theta )\] \[\frac{dy}{d\theta }=a\cos \theta \] \[\Rightarrow \]               \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=-\frac{\cos \theta }{\sin \theta }\] \[\therefore \]Equation of normal at\[[a(1+\cos \theta ),a\sin \theta ]\]is \[(y-a\sin \theta )=\frac{\sin \theta }{\cos \theta }[x-a(1+\cos \theta )]\] It is clear that, in the given options, normal passes through the point (a, 0).

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