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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009

  • question_answer
        If\[\tan {{\theta }_{1}},\tan {{\theta }_{2}}=-\frac{{{a}^{2}}}{{{b}^{2}}},\] then the chord joining two points\[{{\theta }_{1}}\]and\[{{\theta }_{2}}\]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 axis

    D)  end of the minor axis

    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\[OP=\frac{b}{a}\tan {{\theta }_{1}}\] and        \[{{m}_{2}}=\]slope of \[OQ=\frac{b}{a}\tan {{\theta }_{2}}\] \[\therefore \]  \[{{m}_{1}}{{m}_{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\tan {{\theta }_{1}}\tan {{\theta }_{2}}\]                 \[=\frac{{{b}^{2}}}{{{a}^{2}}}\times \frac{-{{a}^{2}}}{{{b}^{2}}}\]                 \[\left[ \therefore \tan {{\theta }_{1}}\tan {{\theta }_{2}}=-\frac{{{a}^{2}}}{{{b}^{2}}}(given) \right]\]                 \[=-1\] \[\therefore \]  \[\angle POQ=\frac{\pi }{2}\] Hence, PQ makes a right angle at the centre of the ellipse.


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