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11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer
    C the centre of the hyperbola\[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. The tangents at any point P on this hyperbola meets the straight lines \[bx-ay=0\]and \[bx+ay=0\] in the points Q and R respectively. Then \[CQ\ .\ CR=\]

    A)            \[{{a}^{2}}+{{b}^{2}}\]              

    B)            \[{{a}^{2}}-{{b}^{2}}\]

    C)            \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]                                   

    D)            \[\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]

    Correct Answer: A

    Solution :

               \[P\]is \[(a\sec \theta ,b\tan \theta )\]            Tangen t at P is \[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\]            It meets \[bx-ay=0\]i.e., \[\frac{x}{a}=\frac{y}{b}\]in Q            \ Q is \[\left( \frac{a}{\sec \theta -\tan \theta },\frac{-b}{\sec \theta -\tan \theta } \right)\]            It meets \[bx+ay=0\] i.e., \[\frac{x}{a}=-\frac{y}{b}\]in R.            \ R is \[\left( \frac{a}{\sec \theta +\tan \theta },\frac{-b}{\sec \theta +\tan \theta } \right)\]            \ \[CQ.CR=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{(\sec \theta -\tan \theta )}.\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{(\sec \theta +\tan \theta )}\]                                  \[={{a}^{2}}+{{b}^{2}}\],  \[\{\because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\}\].


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