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KVPY Sample Paper KVPY Stream-SX Model Paper-24

  • question_answer
    A point P lies on the hyperbola \[9{{x}^{2}}-16{{y}^{2}}=144\] such that \[P{{S}_{1}}\text{ }:\text{ }P{{S}_{2}}\text{ }=\text{ }3/2\] (where \[{{S}_{1}}\] and \[{{S}_{2}}\] are foci of hyperbola). Coordinates of point P is in the first quadrant are-

    A) \[(4\sqrt{2},\,\,3)\]

    B) \[(8,\,\,3\sqrt{3})\]

    C) \[\left( 5,\frac{9}{4} \right)\]

    D) \[(16,\,\,3\sqrt{15})\]

    Correct Answer: D

    Solution :

    \[\frac{{{x}^{2}}}{16}-\frac{{{y}^{2}}}{9}=1\,\,(4\,\sec \theta ,3\tan \theta )\]
    \[\frac{3}{2}\,\,=\,\,\frac{P{{S}_{1}}}{P{{S}_{2}}}\,\,=\left| \frac{\frac{4e\sec \theta +4}{e}}{\frac{4e\sec \theta -4}{e}} \right|=\left| \frac{e\sec \theta +1}{e\sec \theta -1} \right|\]\[\Rightarrow \sec \theta =4\]\[and\,\,\tan \theta =\sqrt{16-1}=\sqrt{15}\]


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