11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer If \[\theta \] and \[\varphi \] are eccentric angles of the ends of a pair of conjugate diameters of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], then \[\theta -\varphi \] is equal to

    A)            \[\pm \frac{\pi }{2}\]               

    B)            \[\pm \pi \]

    C)            0     

    D)            None of thesew

    Correct Answer: A

    Solution :

               Let \[y={{m}_{1}}x\]and \[y={{m}_{2}}x\] be a pair of conjugate diameters of an ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and let \[P(a\cos \theta ,\,b\sin \theta )\] and \[Q(a\cos \varphi ,\,b\sin \varphi )\] be ends of these two diameters. Then \[{{m}_{1}}{{m}_{2}}=-\frac{{{b}^{2}}}{{{a}^{2}}}\]                   \[\Rightarrow \frac{b\sin \theta -0}{a\cos \theta -0}\times \frac{b\sin \varphi -0}{a\cos \varphi -0}=-\frac{{{b}^{2}}}{{{a}^{2}}}\]                    Þ\[\sin \theta \sin \varphi =-\cos \theta \cos \varphi \]Þ\[\cos (\theta -\varphi )=0\Rightarrow \theta -\varphi =\pm \frac{\pi }{2}\].


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