Conic Sections Question 42
Question: C the centre of the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ . The tangents at any point P on this hyperbola meets the straight lines $ bx-ay=0 $ and $ bx+ay=0 $ in the points Q and R respectively. Then $ CQ\ .\ CR= $
Options:
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}} $
Show Answer
Answer:
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 $ ( \frac{a}{\sec \theta -\tan \theta },\frac{-b}{\sec \theta -\tan \theta } ) $
It meets $ bx+ay=0 $ i.e., $ \frac{x}{a}=-\frac{y}{b} $ in R. R is $ ( \frac{a}{\sec \theta +\tan \theta },\frac{-b}{\sec \theta +\tan \theta } ) $
$ 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} $ .
BETA
