Conic Sections Question 535
Question: If PQ is a double ordinate of hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ such that OPQ is an equilateral triangle, O being the centre of the hyperbola. Then the eccentricity e of the hyperbola satisfies
[EAMCET 1999]
Options:
A) $ 1<e<2/\sqrt{3} $
B) $ e=2/\sqrt{3} $
C) $ e=\sqrt{3}/2 $
D) $ e>2/\sqrt{3} $
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Answer:
Correct Answer: D
Solution:
Let P $ (a\sec \theta ,b\tan \theta );Q(a\sec \theta ,-b\tan \theta ) $ be end points of double ordinates and $ C(0,0) $ , is the centre of the hyperbola. Now $ PQ=2b\tan \theta $
$ CQ=CP=\sqrt{a^{2}{{\sec }^{2}}\theta +b^{2}{{\tan }^{2}}\theta } $
Since $ CQ=CP=PQ $ ,
$ \therefore 4b^{2}{{\tan }^{2}}\theta =a^{2}{{\sec }^{2}}\theta +b^{2}{{\tan }^{2}}\theta $
therefore $ 3b^{2}{{\tan }^{2}}\theta =a^{2}{{\sec }^{2}}\theta $
therefore $ 3b^{2}{{\sin }^{2}}\theta =a^{2} $
therefore $ 3a^{2}(e^{2}-1){{\sin }^{2}}\theta =a^{2} $
therefore $ 3(e^{2}-1){{\sin }^{2}}\theta =1 $
therefore $ \frac{1}{3(e^{2}-1)}={{\sin }^{2}}\theta <1 $ , $ (\because {{\sin }^{2}}\theta <1) $
$ \Rightarrow $ $ \frac{1}{e^{2}-1}<3 $
$ \Rightarrow e^{2}-1>\frac{1}{3} $
$ \Rightarrow e^{2}>\frac{4}{3} $
$ \Rightarrow e>\frac{2}{\sqrt{3}} $ .
BETA
