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JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    The distance of the point \['\theta '\]on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] from a focus is

    A)            \[a(e+\cos \theta )\]                

    B)            \[a(e-\cos \theta )\]

    C)            \[a(1+e\cos \theta )\]              

    D)            \[a(1+2e\cos \theta )\]

    Correct Answer: C

    Solution :

               Focal distance of any point P (x,y) on the ellipse is equal to \[SP=a+ex\]. Here \[x=a\cos \theta \]            Here\[SP=a+ae\cos \theta =a(1+e\cos \theta )\].


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