Conic-Sections Question 572
Question: Let $ S_1,S_2 $ be the foci of the ellipse, $ \frac{x^{2}}{16}+\frac{y^{2}}{8}=1 $ . If $ A(x+y) $ is any point on the ellipse, then the maximum area of the triangle $ AS_1S_2 $ (in square units) is
Options:
A) $ 2\sqrt{2} $
B) $ 2\sqrt{3} $
C) 8
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Equation of ellipse is $ \frac{x^{2}}{16}+\frac{y^{2}}{8}=1 $ where, $ a=4,b=2\sqrt{2} $ Eccentricity, $ e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{8}{16}}=\frac{1}{\sqrt{2}} $ Area is maximum when vertex is (0, b)
$ \therefore $ Maximum area $ =\frac{1}{2}\times 2ae\times b $ $ =\frac{1}{2}\times 2\times 4\times 2\sqrt{2}\times \frac{1}{\sqrt{2}}=8 $ sq. units
BETA
