Conic Sections Self Evaluation Test Conic Sections Question 66
Question: If $ P\equiv (x,y),F_1\equiv (3,0),F_2\equiv (-3,0) $ and $ 16x^{2}+25y^{2}=400, $ then $ PF_1+PF_2 $ equals
Options:
A) 8
B) 6
C) 10
D) 12
Answer:
Correct Answer: C
Solution:
[c] The ellipse can be written as, $ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1 $ Here $ a^{2}=25,b^{2}=16, $ but $ b^{2}=a^{2}(1-e^{2})\Rightarrow 16/25 $ $ =1-e^{2} $
$ \Rightarrow e^{2}=1-16/25=9/25\Rightarrow e=3/5 $ Foci of the ellipse are $ (\pm ae,0)=(\pm 3,0), $ i.e., $ F_1 $ and $ F_2 $
$ \therefore $ We have $ PF_1+PF_2=2a=10 $ for every point P on the ellipse.
BETA
