Conic Sections Question 442
Question: The eccentricity of the conjugate hyperbola of the hyperbola $ x^{2}-3y^{2}=1 $ , is
[MP PET 1999]
Options:
A) 2
B) $ \frac{2}{\sqrt{3}} $
C) 4
D) $ \frac{4}{3} $
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Answer:
Correct Answer: A
Solution:
Eccentricity of $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ is $ e=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}} $
Eccentricity of conjugate hyperbola, $ e’=\sqrt{\frac{a^{2}+b^{2}}{b^{2}}} $
Write the given equation in standard form, $ \frac{x^{2}}{1}-\frac{y^{2}}{1/3}=1 $
therefore $ a^{2}=1,b^{2}=\frac{1}{3} $
$ e’=\sqrt{\frac{1+1/3}{1/3}}=\sqrt{4}=2 $ .
BETA
