CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The point on the curve\[{{x}^{2}}+{{y}^{2}}={{a}^{2}},\text{ }y\ge 0\]at which the tangent is parallel to\[x-\]axis is

    A)  \[(a,0)\]                             

    B)  \[(-a,0)\]

    C)  \[\left( \frac{a}{2},\frac{\sqrt{3}}{2}a \right)\]  

    D)         \[(0,a)\]

    E)  \[(0,{{a}^{2}})\]

    Correct Answer: D

    Solution :

    Given curve is \[{{x}^{2}}+{{y}^{2}}={{a}^{2}},y\ge 0\] ?. (i) \[\Rightarrow \]               \[2x+2y\frac{dy}{dx}=0\] (Differentiating w.r.t.\[x\]on both sides) \[\Rightarrow \]               \[\frac{dy}{dx}=-\frac{x}{y}\] \[\because \]The tangent is parallel to\[x-\]axis. \[\therefore \]  \[\frac{dy}{dx}=0\] \[\Rightarrow \]               \[x=0\] Putting\[x=0\]in Eq. (i), we get                 \[{{y}^{2}}={{a}^{2}}\] \[\Rightarrow \]               \[y=\pm a\] \[\therefore \]Required point \[=(0,\text{ }a)\]


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