JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    The eccentric angles of the extremities of latus recta of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] are given by

    A)            \[{{\tan }^{-1}}\left( \pm \frac{ae}{b} \right)\]                          

    B)            \[{{\tan }^{-1}}\left( \pm \frac{be}{a} \right)\]

    C)            \[{{\tan }^{-1}}\left( \pm \frac{b}{ae} \right)\]                          

    D)            \[{{\tan }^{-1}}\left( \pm \frac{a}{be} \right)\]

    Correct Answer: C

    Solution :

               Coordinates of any point on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] whose eccentric angle is \[\theta \]are \[(a\cos \theta ,\,\,b\sin \theta ).\]            The coordinates of the end points of latus recta are \[\left( ae,\,\pm \frac{{{b}^{2}}}{a} \right).\] \[\therefore a\cos \theta =ae\] and \[b\sin \theta =\pm \frac{{{b}^{2}}}{a}\]                    Þ \[\tan \theta =\pm \frac{b}{ae}\Rightarrow \theta ={{\tan }^{-1}}\left( \pm \frac{b}{ae} \right)\].


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