Conic Sections Question 265
Question: The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is:
Options:
A) $ \frac{4}{3} $
B) $ \frac{4}{\sqrt{3}} $
C) $ \frac{2}{\sqrt{3}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The standard equation of hyperbola is $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ Latus rectum $ =\frac{2b^{2}}{a}, $ Conjugate axis $ =2b $ , Distance between the foci =2ae According to the question, $ \frac{2b^{2}}{a}=8 $
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(i) $ 2b=\frac{1}{2}(2ae)\Rightarrow b=\frac{ae}{2} $
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(ii) From (i) & (ii), $ \frac{2}{a}{{( \frac{ae}{2} )}^{2}}=8 $
$ \Rightarrow 2.\frac{a^{2}e^{2}}{4a}=8 $
$ \Rightarrow ae^{2}=16 $
- (iii) From (i), $ b^{2}=4a $ Using $ b^{2}=a^{2}(e^{2}-1) $ we get $ (4a)=a^{2}(e^{2}-1)\Rightarrow 4=\frac{16}{e^{2}}(e^{2}-1) $
$ \Rightarrow 16-\frac{16}{e^{2}}=4\Rightarrow \frac{16}{e^{2}}=12\therefore e=\frac{2}{\sqrt{3}}. $
BETA
