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BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    If focii of \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] coincide with the foci of \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\]and eccentricity of the  hyperbola is 2, then

    A)  \[{{a}^{2}}+{{b}^{2}}=14\]

    B)  there is a director circle of the hyperbola

    C)  centre of the director circle is (0, 0)

    D)  length of latusrectum of the hyperbola is

    Correct Answer: D

    Solution :

    For the ellipse, a = 5 and \[e=\sqrt{\frac{25-9}{25}}=\frac{4}{5}\] \[\therefore \]  \[ae=4\] Hence, the focii are (-4, 0) and (4, 0). For the hyperbola, ae = 4, e = 2 \[\therefore \]  a = 2 \[{{b}^{2}}=4(4-1)=12\] \[b=\sqrt{12}\] Length of latusrectum \[=\frac{2{{b}^{2}}}{a}=2\times \frac{12}{2}\]                                                 \[=12\]


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