A) \[2\sqrt{2}\]
B) \[2\sqrt{3}\]
C) 8
D) 4
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
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