Conic Sections Question 473
Question: The eccentricity of the hyperbola conjugate to $ x^{2}-3y^{2}=2x+8 $ is
[UPSEAT 2004]
Options:
A) $ \frac{2}{\sqrt{3}} $
B) $ \sqrt{3} $
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given, equation of hyperbola is $ x^{2}-3y^{2}=2x+8 $
therefore $ x^{2}-2x-3y^{2}=8 $
therefore $ {{(x-1)}^{2}}-3y^{2}=9 $
therefore $ \frac{{{(x-1)}^{2}}}{9}-\frac{y^{2}}{3}=1 $
Conjugate of this hyperbola is $ -\frac{{{(x-1)}^{2}}}{9}+\frac{y^{2}}{3}=1 $
and its eccentricity $ (e)=\sqrt{( \frac{a^{2}+b^{2}}{b^{2}} )} $
Here, $ a^{2}=9 $ , $ b^{2}=3 $ ;
$ \therefore $ $ e=\sqrt{\frac{9+3}{3}}=2 $ .
BETA
