A) \[{{a}^{2}}+{{b}^{2}}\]
B) \[{{a}^{2}}-{{b}^{2}}\]
C) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]
D) \[\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]
Correct Answer: A
Solution :
\[P\]is \[(a\sec \theta ,b\tan \theta )\] Tangen t at P is \[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\] It meets \[bx-ay=0\]i.e., \[\frac{x}{a}=\frac{y}{b}\]in Q \ Q is \[\left( \frac{a}{\sec \theta -\tan \theta },\frac{-b}{\sec \theta -\tan \theta } \right)\] It meets \[bx+ay=0\] i.e., \[\frac{x}{a}=-\frac{y}{b}\]in R. \ R is \[\left( \frac{a}{\sec \theta +\tan \theta },\frac{-b}{\sec \theta +\tan \theta } \right)\] \ \[CQ.CR=\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{(\sec \theta -\tan \theta )}.\frac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{(\sec \theta +\tan \theta )}\] \[={{a}^{2}}+{{b}^{2}}\], \[\{\because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\}\].You need to login to perform this action.
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