Conic Sections Question 236
Question: If AB is a double ordinate of the hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ such that $ \Delta OAB $ is an equilateral triangle O being the origin, then the eccentricity of the hyperbola satisfies.
Options:
A) $ e>\sqrt{3} $
B) $ 1<e<\frac{2}{\sqrt{3}} $
C) $ e=\frac{2}{\sqrt{3}} $
D) $ e>\frac{2}{\sqrt{3}} $
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Answer:
Correct Answer: D
Solution:
[d] Let the length of the double ordinate be $ 2\ell . $
$ \therefore AB=2\ell $ and $ AM=BM=\ell $ Clearly ordinate of point A is $ \ell $ . The abscissa of the point A is given by $ \frac{x^{2}}{a^{2}}-\frac{{{\ell }^{2}}}{b^{2}}=1\Rightarrow x=\frac{a\sqrt{b^{2}+{{\ell }^{2}}}}{b} $
$ \therefore $ A is $ ( \frac{a\sqrt{b^{2}+{{\ell }^{2}}}}{b},\ell ) $ Since $ \Delta OAB $ is equilateral triangle, therefore $ OA=AB=OB=2\ell . $ Also, $ OM^{2}+AM^{2}=OA^{2} $
$ \therefore \frac{a^{2}(b^{2}+{{\ell }^{2}})}{b^{2}}+{{\ell }^{2}}=4{{\ell }^{2}} $ We get $ {{\ell }^{2}}=\frac{a^{2}b^{2}}{3b^{2}-a^{2}} $ Since $ {{\ell }^{2}}>0\therefore \frac{a^{2}b^{2}}{3b^{2}-a^{2}}>0\Rightarrow 3b^{2}-a^{2}>0 $
$ \Rightarrow 3a^{2}(e^{2}-1)-a^{2}>0\Rightarrow e>\frac{2}{\sqrt{3}} $
BETA
