Conic Sections Question 482
Question: The equation of the hyperbola in the standard form (with transverse axis along the x-axis) having the length of the latus rectum = 9 units and eccentricity = 5/4 is
[Kerala (Engg.) 2005]
Options:
A) $ \frac{x^{2}}{16}-\frac{y^{2}}{18}=1 $
B) $ \frac{x^{2}}{36}-\frac{y^{2}}{27}=1 $
C) $ \frac{x^{2}}{64}-\frac{y^{2}}{36}=1 $
D) $ \frac{x^{2}}{36}-\frac{y^{2}}{64}=1 $
E) $ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \because \frac{2b^{2}}{a^{2}}=9 $
therefore $ 2b^{2}=9a $ ……(i) Now $ b^{2}=a^{2}(e^{2}-1)=\frac{9}{16}a^{2} $
therefore $ a=\frac{4}{3}b $ …..(ii), ( $ \because $
$ e=\frac{5}{4} $ ) From (i) and (ii), $ b=6 $ , $ a=8 $
Hence, equation of hyperbola $ \frac{x^{2}}{64}-\frac{y^{2}}{36}=1 $ .
BETA
