• # 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)            $-\left( \frac{{{a}^{2}}+{{b}^{2}}}{b} \right)$

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 $\tan \varphi =\cot \theta$            $\therefore y=k=\frac{{{a}^{2}}+{{b}^{2}}}{b}.\frac{\tan \theta \cos \theta -\cot \theta \sin \theta }{\cos \theta -\sin \theta }$                    $=\frac{{{a}^{2}}+{{b}^{2}}}{b}\left( \frac{\sin \theta -\cos \theta }{\cos \theta -\sin \theta } \right)=-\frac{({{a}^{2}}+{{b}^{2}})}{b}$.