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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