CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    On the ellipse\[4{{x}^{2}}+9{{y}^{2}}=1\]the point at which the tangent are parallel to\[8x=9y\]are:

    A)  \[\left( \frac{2}{5},\frac{1}{5} \right)or\left( -\frac{2}{5},-\frac{1}{5} \right)\]

    B)  \[\left( -\frac{2}{5},\frac{1}{5} \right)or\left( \frac{2}{5},-\frac{1}{5} \right)\]

    C)  \[\left( -\frac{2}{5},-\frac{1}{5} \right)\]

    D)  \[\left( -\frac{3}{5},-\frac{2}{5} \right)or\left( \frac{3}{5},\frac{2}{5} \right)\]

    E)  \[\left( -\frac{3}{5},\frac{2}{5} \right)or\left( \frac{3}{5},-\frac{2}{5} \right)\]

    Correct Answer: B

    Solution :

    We have, \[{{a}^{2}}=\frac{1}{4},{{b}^{2}}=\frac{1}{9},m=\frac{8}{9}\] The required points are \[\left( \pm \frac{{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}},\mp \frac{{{b}^{2}}}{\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}} \right)\] \[\left( \pm \frac{\frac{1}{4}.\frac{8}{9}}{\sqrt{\frac{1}{4}\times \frac{64}{84}+\frac{1}{9}}},\mp \frac{\frac{1}{9}}{\sqrt{\frac{1}{4}\times \frac{64}{81}+\frac{1}{9}}} \right)\] i.e.,        \[\left( \pm \frac{2}{5}\mp \frac{1}{5} \right)\]


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