Conic Sections Question 443
Question: If e and e- are eccentricities of hyperbola and its conjugate respectively, then
[UPSEAT 1999]
Options:
A) $ {{( \frac{1}{e} )}^{2}}+{{( \frac{1}{e’} )}^{2}}=1 $
B) $ \frac{1}{e}+\frac{1}{e’}=1 $
C) $ {{( \frac{1}{e} )}^{2}}+{{( \frac{1}{e’} )}^{2}}=0 $
D) $ \frac{1}{e}+\frac{1}{e’}=2 $
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Answer:
Correct Answer: A
Solution:
Let hyperbola is $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ ……(i) Then its conjugate will be, $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1 $ …..(ii) If $ e $ is eccentricity of hyperbola (i), then $ b^{2}=a^{2}(e^{2}-1) $
or $ \frac{1}{e^{2}}=\frac{a^{2}}{(a^{2}+b^{2})} $ ……(iii) Similarly if e’ is eccentricity of conjugate (ii), then $ a^{2}=b^{2}(e{{’}^{2}}-1) $ or $ \frac{1}{e{{’}^{2}}}=\frac{b^{2}}{(a^{2}+b^{2})} $ ……(iv)
Adding (iii) and (iv), $ \frac{1}{{{(e’)}^{2}}}+\frac{1}{e^{2}}=\frac{a^{2}}{a^{2}+b^{2}}+\frac{b^{2}}{a^{2}+b^{2}}=1. $
BETA
