Conic Sections Question 383
Question: In an ellipse $ 9x^{2}+5y^{2}=45 $ , the distance between the foci is
[Karnataka CET 2002]
Options:
A) $ 4\sqrt{5} $
B) $ \frac{49}{4}x^{2}-\frac{51}{196}y^{2}=1 $
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Show Answer
Answer:
Correct Answer: D
Solution:
Given equation may be written as $ \frac{x^{2}}{5}+\frac{y^{2}}{9}=1. $
Comparing the given equation with standard equation, we get $ a^{2}=5 $ and $ b^{2}=9. $ We also know that in an ellipse (where $ a^{2}<b^{2}) $
$ a^{2}=b^{2}(1-e^{2})\lambda $ or $ e^{2}=\frac{b^{2}-a^{2}}{b^{2}}=\frac{9-5}{9}=\frac{4}{9} $ or $ e=\frac{2}{3}. $
Therefore distance between foci $ =2ae=2\times 3\times \frac{2}{3}=4 $ .
BETA
