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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    Suppose S and S are foci of the ellipse\[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1.\]If p is a variable point on the ellipse and if \[\Delta \] is area of the triangle PSS then the maximum value of\[\Delta \]is:

    A)  8                                            

    B)  12

    C)  16                         

    D)         20

    E)  24

    Correct Answer: B

    Solution :

    We have,\[{{a}^{2}}=25\]and\[{{b}^{2}}=16\] \[\therefore \]\[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\]    So, the co-ordinates of foci S and S are (3, 0) and\[(-3,0)\]respectively. Let\[P(5\cos \theta ,4\sin \theta )\] be a variable point on the ellipse. Then\[\Delta =area\text{ }of\Delta \text{ }PSS=12\text{ }sin\theta .\] So maximum value of area of\[\Delta PSS\]is 12, since value of\[\sin \theta \]lies between\[-1\]and 1.


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