11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer Let \[P(a\sec \theta ,\ b\tan \theta )\] and \[Q(a\sec \varphi ,\ b\tan \varphi )\], where \[\theta +\varphi =\frac{\pi }{2}\], be two points on the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. If (h, k) is the point of intersection of the normals at P and Q, then k is equal to                                                                            [IIT 1999;  MP PET 2002]

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

    B)            \[-\left( \frac{{{a}^{2}}+{{b}^{2}}}{a} \right)\]

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

    D)            \[-\left( \frac{{{a}^{2}}+{{b}^{2}}}{b} \right)\]

    Correct Answer: D

    Solution :

               Given P\[(a\sec \theta ,\,b\tan \theta )\]and  \[Q\,(a\sec \varphi ,\,b\tan \varphi )\]                   The equation of tangent at point P is \[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\]                   m of tangent \[=\frac{b}{\tan \theta }\times \frac{\sec \theta }{a}=\frac{b}{a}.\frac{1}{\sin \theta }\]                   Hence the equation of perpendicular at P is            \[y-b\tan \theta =-\frac{a\sin \theta }{b}(x-a\sec \theta )\]            or \[by-{{b}^{2}}\tan \theta =-a\sin \theta \,x+{{a}^{2}}\tan \theta \]            or \[a\sin \theta \,x+by=({{a}^{2}}+{{b}^{2}})\tan \theta \]                              .....(i)            Similarly the equation of perpendicular at Q is            \[a\sin \varphi \,x+by=({{a}^{2}}+{{b}^{2}})\tan \varphi \]                                ?..(ii)            On multiplying (i) by \[\sin \varphi \]and (ii) by \[\sin \theta \]            \[a\sin \theta \sin \varphi \,x+b\sin \varphi y=({{a}^{2}}+{{b}^{2}})\tan \theta \sin \varphi \]            \[a\sin \varphi \sin \theta \,x+b\sin \theta y=({{a}^{2}}+{{b}^{2}})\tan \varphi \sin \theta \]            On subtraction by,            \[(\sin \varphi -\sin \theta )=({{a}^{2}}+{{b}^{2}})(\tan \theta \sin \varphi -\tan \varphi \sin \theta )\]            \[\therefore y=k=\frac{{{a}^{2}}+{{b}^{2}}}{b}.\frac{\tan \theta \sin \varphi -\tan \varphi \sin \theta }{\sin \varphi -\sin \theta }\]            \[\because \theta +\varphi =\frac{\pi }{2}\] Þ \[\varphi =\frac{\pi }{2}-\theta \]            Þ\[\sin \varphi =\cos \theta \] and \[\tan \varphi =\cot \theta \]            \[\therefore y=k=\frac{{{a}^{2}}+{{b}^{2}}}{b}.\frac{\tan \theta \cos \theta -\cot \theta \sin \theta }{\cos \theta -\sin \theta }\]                    \[=\frac{{{a}^{2}}+{{b}^{2}}}{b}\left( \frac{\sin \theta -\cos \theta }{\cos \theta -\sin \theta } \right)=-\frac{({{a}^{2}}+{{b}^{2}})}{b}\].

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