CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The distance between the foci of the conic \[7{{x}^{2}}-9{{y}^{2}}=63\]is equal to

    A)  8         

    B)  4

    C)  3                            

    D)         7

    E)  12

    Correct Answer: A

    Solution :

       Equation of given conic is \[7{{x}^{2}}-9{{y}^{2}}=63\] \[\Rightarrow \]               \[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{7}=1\] Comparing it with hyperbola                 \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \]               \[a=3,b=\sqrt{7}\] \[\Rightarrow \]               \[a>b\] \[\Rightarrow \]               \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\]                 \[=\sqrt{1+\frac{7}{9}}=\frac{4}{3}\] \[\therefore \]Foci = (4, 0) and \[(-4,0)\] \[\therefore \]Required distance\[=\sqrt{{{(4+4)}^{2}}}=8\]                      


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