CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    The point on the curve \[{{y}^{2}}=x,\]the tangent at which makes an angle \[{{45}^{o}}\] with -X-axis is

    A)  \[\left( \frac{1}{4},\frac{1}{2} \right)\]                  

    B)  \[\left( \frac{1}{2},\frac{1}{4} \right)\]

    C)  \[\left( \frac{1}{2},\frac{-1}{2} \right)\]                

    D)  \[\left( \frac{1}{2},\frac{1}{2} \right)\]

    Correct Answer: A

    Solution :

    Given,            \[{{y}^{2}}=x\]              ...(i)    \[\Rightarrow \]                               \[2y\frac{dy}{dx}=1\] \[\Rightarrow \]                               \[\frac{dy}{dx}=\frac{1}{2y}=slope\] Also given,          \[\theta ={{45}^{o}}\] \[\therefore \]  Slope tan \[\tan \,{{45}^{o}}=1\] \[\Rightarrow \]               \[\frac{1}{2y}=1\] \[\Rightarrow \]               \[y=\frac{1}{2}\] From Eq. (i), if  \[y=1/2,\] then \[x=\frac{1}{4}\] \[\therefore \]  Required point is \[\left( \frac{1}{4},\frac{1}{2} \right),\]


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