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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    If the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] intersects the hyperbola \[xy={{c}^{2}}\]four points \[P({{x}_{1}},{{y}_{1}}),\] \[Q({{x}_{2}},{{y}_{2}}),\] \[R({{x}_{3}},{{y}_{3}})\] and \[S({{x}_{4}},{{y}_{4}})\] then,

    A)  \[{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}=2\]

    B)  \[{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}=2{{c}^{4}}\]

    C)  \[{{y}_{1}}{{y}_{2}}{{y}_{3}}{{y}_{4}}=2{{c}^{4}}\]

    D)  \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0\]

    Correct Answer: D

    Solution :

    Given,    \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] and \[xy={{c}^{2}}\] \[\therefore \]  \[{{x}^{2}}+\left( \frac{{{c}^{2}}}{x} \right)={{a}^{2}}\] \[\Rightarrow \]               \[{{x}^{4}}-{{a}^{2}}{{x}^{2}}+{{c}^{4}}=0\] \[\therefore \]  \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0\]


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