A) \[PG=P{{G}^{2}}\]
B) \[PG=PC\]
C) \[PG=Gg\]
D) \[PC=Cg\]
Correct Answer: B
Solution :
Let the equation of the rectangular hyperbola be \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}\]. Let P be any point on the given rectangular hyperbola specified as (a \[\sec \,\theta \], a \[\tan \,\theta \]), then the equation of the normal to the hyperbola at P is \[x\,\sin \,\theta +y=2a\,\tan \,\theta \] which will intersect axes at \[G(2\,a\,\sec \,\theta ,0)\] and \[g(0,\,2a\,\tan \,\theta )\] Now, \[P{{G}^{2}}={{(a\,\sec \,\theta -2a\,\sec \,\theta )}^{2}}+{{(a\,\tan \,\theta -0)}^{2}}\] \[={{a}^{2}}({{\sec }^{2}}\theta +{{\tan }^{2}}\theta )\] \[P{{g}^{2}}={{(a\,\sec \,\theta -0)}^{2}}+{{(a\,\tan \,\theta -2a\,\tan \,\theta )}^{2}}\] \[={{a}^{2}}({{\sec }^{2}}\theta +{{\tan }^{2}}\theta )\] and \[P{{C}^{2}}={{(a\,\sec \,\theta -0)}^{2}}+{{(a\,\tan \,\theta \,-0)}^{2}}\] \[={{a}^{2}}({{\sec }^{2}}\theta +{{\tan }^{2}}\theta )\] Clearly, \[PG=Pg=PC\]You need to login to perform this action.
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