JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    The curve that passes through the point (2, 3), and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact is given by:     AIEEE  Solved  Paper (Held On 11 May  2011)

    A) \[2y-3x=0\]

    B) \[y=\frac{6}{x}\]

    C) \[{{x}^{2}}+{{y}^{2}}=13\]

    D) \[{{\left( \frac{x}{2} \right)}^{2}}+{{\left( \frac{y}{3} \right)}^{2}}=2\]

    Correct Answer: B

    Solution :

                    \[Y-y=\frac{dy}{dx}(X-x)\]                 X-intercept is \[\left( x-\frac{y}{dy/dx},0 \right)\]                 Y=intercept is \[\left( 0,y-\frac{xdy}{dx} \right)\]                 According to statement                 \[x-\frac{y}{dy/dx}=2x\]and\[y-\frac{xdy}{dx}=2y\]                 \[\frac{\frac{-y}{dy}}{dx}=x\]                     \[\frac{-xdy}{dx}=y\]                 \[\frac{dx}{x}+\frac{dy}{y}=0\]  \[\ell ny=-\ell nx+\ell nc\] \[y=\frac{c}{x}\Rightarrow c=6\] Hence\[y=\frac{6}{x}\]                

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