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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

  • question_answer
    If \[\tan \,\,{{q}_{1}}\]\[\tan \,\,{{q}_{2}}=-\frac{{{a}^{2}}}{{{b}^{2}}},\] then the chord joining two points \[(a\,\,\cos \,\,{{q}_{1}},\,\,b\,\,\sin \,\,{{q}_{1}}),\] and \[(a\,\,\cos \,\,{{q}_{2}},\,\,b\,\,\sin \,\,{{q}_{2}})\] on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] will subtend a right angle at

    A) Focus              

    B) Centre

    C) End of the major axes             

    D) End of minor axes

    Correct Answer: B

    Solution :

    Let \[P(a\,\,\cos {{\theta }_{1}},\,b\sin {{\theta }_{1}})\] and \[Q\,(a\,\cos {{\theta }_{2}},\,b\sin {{\theta }_{2}})\]be two points on the ellipse. Then
    \[{{m}_{1}}=\] Slope of OP (O is centre) \[=\frac{b}{a}\,\tan {{\theta }_{1}};\]
    \[{{m}_{2}}=\] Slope of \[OQ=\frac{b}{a}\tan {{\theta }_{2}}\]
    Since,  \[{{m}_{1}}{{m}_{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\,\tan {{\theta }_{1}}\,\tan {{\theta }_{2}}\]
    \[=\frac{{{b}^{2}}}{{{a}^{2}}}\left( -\frac{{{a}^{2}}}{{{b}^{2}}} \right)=-1\,\,(\because \,\,\tan {{\theta }_{1}}\,\tan {{\theta }_{2}}=\frac{-{{a}^{2}}}{{{b}^{2}}})\]
    \[\therefore \,\,\angle POQ=90{}^\circ .\] Hence PQ makes a right angle at the centre of the ellipse.


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