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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
              Let C be a curve given by\[y(x)=1+\sqrt{4x-3},x>\frac{3}{4}\]. If P is a point on C, such that the tangent at P has slope\[\frac{2}{3}\], then a point through which the normal at P passes, is   JEE Main Online Paper (Held On 10 April 2016)

    A) (3, - 4)                     

    B) (1, 7)

    C) (4, - 3)                     

    D) (2, 3)

    Correct Answer: B

    Solution :

                              \[y(x)=1+\sqrt{4x-3},x>\frac{3}{4}\]              Let P\[(\alpha ,\,1+\left( \sqrt{4\alpha -3} \right)\] be the point.              at which \[\frac{dy}{dx}ATP=\frac{2}{3}\] \[\Rightarrow \frac{2}{\sqrt{4\alpha -3}}=\frac{2}{3}\] \[\Rightarrow 4\alpha -3=9\] \[\Rightarrow \alpha =3\]                                                Hence P(3,4) slope of normal at P(3,4) is \[=-\frac{3}{2}\] equation of normal \[Y-4=-\frac{3}{2}(X-3)\] \[2y-8=3x+9\] \[3x+2y=17\] clearly it is passes through (1,7)


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