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JEE Main & Advanced Mathematics Conic Sections Question Bank Self Evaluation Test - Conic Sections

  • question_answer
    The sum of the focal distances of  a point on the ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\] is:

    A) 4 units

    B) 6 units

    C) 8 units

    D) 10 units

    Correct Answer: A

    Solution :

    [a] Given equation of ellipse is \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\] \[\Rightarrow \frac{{{x}^{2}}}{{{(2)}^{2}}}+\frac{{{y}^{2}}}{{{(3)}^{2}}}=1\Rightarrow a=2\] and \[b=3\] Length of major axis \[=2a=4\] Since, we have Sum of the focal distance of a point on ellipse = length of major axis. \[\therefore \] Required Ans = 4 units.


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