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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    The foci of Ac ellipse\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]and the hyperbola\[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\]coincide. Then, the value of\[{{b}^{2}}\]is:

    A)  1                                            

    B)  5

    C)  7                            

    D)         9            

    E)  36

    Correct Answer: C

    Solution :

    Foci of ellipse\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[(\pm \,4e,\,0)\] and foci of hyperbola \[\frac{{{x}^{2}}}{\frac{144}{25}}-\frac{{{y}^{2}}}{\frac{81}{85}}=1\]is\[\left( \pm \frac{12}{5}e,0 \right)\] According to the given condition\[4e=12\text{ }e\] \[4\sqrt{1-\frac{{{b}^{2}}}{16}}=\frac{12}{5}\sqrt{1+\frac{81}{144}}\] \[\Rightarrow \]               \[\sqrt{1-\frac{{{b}^{2}}}{16}}=\frac{3}{5}\times \frac{15}{12}=\frac{3}{4}\] \[\Rightarrow \]               \[\frac{{{b}^{2}}}{16}=1-\frac{9}{16}=\frac{7}{16}\] \[\Rightarrow \]               \[{{b}^{2}}=7\]


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