Conic Sections Question 387
Question: In an ellipse the distance between its foci is 6 and its minor axis is 8. Then its eccentricity is
[EAMCET 1994]
Options:
A) $ \frac{4}{5} $
B) $ \frac{1}{\sqrt{52}} $
C) $ \frac{3}{5} $
D) $ 25x^{2}+144y^{2}=900 $
Show Answer
Answer:
Correct Answer: C
Solution:
Distance between foci $ =6 $
therefore $ ae=3 $ . Minor axis $ =8 $
therefore $ 2b=8 $
therefore $ b=4 $
therefore $ b^{2}=16 $
therefore $ a^{2}(1-e^{2})=16 $
therefore $ a^{2}-a^{2}e^{2}=16 $
therefore $ a^{2}-9=16 $
thereforea=5
Hence $ ae=3 $
therefore $ e=\frac{3}{5} $ .
BETA
