Conic Sections Question 13
Question: If the eccentricity of the ellipse $ \frac{x^{2}}{a^{2}+1}+\frac{y^{2}}{a^{2}+2}=1 $ Is $ 1\sqrt{6} $ , then the latus rectum of the ellipse is
Options:
A) $ 5/\sqrt{6} $
B) $ 10/\sqrt{6} $
C) $ 8/\sqrt{6} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Here, $ a^{2}+2>a^{2}+1 $ Or $ a^{2}+1=(a^{2}+2)(1-e^{2}) $ Or $ a^{2}+1=(a^{2}+2)\frac{5}{6} $
Or $ 6a^{2}+6=5a^{2}+10 $ Or $ a^{2}=10-6=4 $
Or $ a=\pm 2 $
Latus rectum $ \frac{2(a^{2}+1)}{\sqrt{a^{2}+2}}=\frac{2\times 5}{\sqrt{6}}=\frac{10}{\sqrt{6}} $
BETA
