Conic Sections Question 380
Question: If the eccentricity of the two ellipse $ \frac{x^{2}}{169}+\frac{y^{2}}{25}=1 $ and $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ are equal, then the value of $ a/b $ is
[UPSEAT 2001]
Options:
A) 5/13
B) 6/13
C) 13/5
D) 13/6
Show Answer
Answer:
Correct Answer: C
Solution:
In the first case, eccentricity $ e=\sqrt{1-(25/169)} $
In the second case, $ e’=\sqrt{1-(b^{2}/a^{2})} $
According to the given condition, $ \sqrt{1-b^{2}/a^{2}}=\sqrt{1-(25/169)} $
$ \Rightarrow b/a=5/13 $ , $ (\because a>0,b>0) $
therefore $ a/b=13/5. $
BETA
