CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    The curves\[4{{x}^{2}}+9{{y}^{2}}=72\]and\[{{x}^{2}}-{{y}^{2}}=5\]at (3, 2):

    A)  touch each other

    B)        cut orthogonally

    C)  intersect at\[45{}^\circ \]

    D)         intersect at\[60{}^\circ \]

    E)  none of these

    Correct Answer: B

    Solution :

    The given curves are \[4{{x}^{2}}+9{{y}^{2}}=72\]                  ...(i) and         \[{{x}^{2}}-{{y}^{2}}=5\]                          ?(ii) On differentiating Eq. (i) w. r. t.\[x,\]we get                 \[8x+18y\frac{dy}{dx}=0\]                 \[\frac{dy}{dx}=-\frac{4x}{9y}\] \[\therefore \]Slope of Eq.(i)\[={{m}_{1}}={{\left( \frac{dy}{dx} \right)}_{(3,2)}}=-\frac{2}{3}\] On differenting Eq. (ii) w. r. t.\[x,\]we get                 \[2x-2y\frac{dy}{dx}=0\] \[\Rightarrow \]               \[\frac{dy}{dx}=\frac{x}{y}\] \[\therefore \]Slope of Eq.(ii) \[={{m}^{2}}={{\left( \frac{dy}{dx} \right)}_{(3,2)}}=\frac{3}{2}\] \[\therefore \]  \[{{m}_{1}}{{m}_{2}}=\frac{-2}{3}\times \frac{3}{2}=-1\] \[\therefore \]Both the curves cut orthogonally.


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