Conic Sections Question 363
Question: The equation of the ellipse whose one focus is at (4, 0) and whose eccentricity is 4/5, is
[Karnataka CET 1993]
Options:
A) $ \frac{x^{2}}{3^{2}}+\frac{y^{2}}{5^{2}}=1 $
B) $ \frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1 $
C) $ \frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1 $
D) $ \frac{x^{2}}{4^{2}}+\frac{y^{2}}{5^{2}}=1 $
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ ae=4 $ and $ e=\frac{4}{5}\Rightarrow a=5 $
Now $ b^{2}=a^{2}(1-e^{2})\Rightarrow b^{2}=25( 1-\frac{16}{25} )=9 $
Hence equation of the ellipse is $ \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 $ .
BETA
