question_answer

Let normal at any point P to the rectangular hyperbola meet the axes in G and g and C be the centre of the hyperbola. Then, which of the following is true

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, \[PG=Pg=PC\]