Conic Sections Question 314
Question: The equation of the ellipse whose foci are $ (\pm 5,\ 0) $ and one of its directrix is $ 5x=36 $ , is
Options:
A) $ \frac{x^{2}}{36}+\frac{y^{2}}{11}=1 $
B) $ \frac{x^{2}}{6}+\frac{y^{2}}{\sqrt{11}}=1 $
C) $ \frac{x^{2}}{6}+\frac{y^{2}}{11}=1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Foci $ (\pm 5,0)\equiv (\pm ae,0) $ . Directrix $ ( x=\frac{36}{5} )\equiv x=\frac{a}{e} $
So, $ \frac{a}{e}=\frac{36}{5},\ ae=5 $
therefore $ a=6 $ and $ e=\frac{5}{6} $
Therefore, $ b=6\sqrt{1-\frac{25}{36}}=6\frac{\sqrt{11}}{6}=\sqrt{11} $
Hence equation is $ \frac{x^{2}}{36}+\frac{y^{2}}{11}=1 $ .
BETA
