A) \[4\,{{a}^{2}}\]sq. unit
B) \[12\,{{a}^{2}}\]sq. unit
C) \[4\,{{a}^{3}}\]sq. unit
D) None of these
Correct Answer: A
Solution :
The parabola meets x-axis at the points, where \[\frac{3}{a}({{a}^{2}}-{{x}^{2}})=0\Rightarrow x=\pm a.\] So the required area \[=\int_{-a}^{a}{\frac{3}{a}({{a}^{2}}-{{x}^{2}})dx=\frac{6}{a}\int_{0}^{a}{\,({{a}^{2}}-{{x}^{2}})dx=4{{a}^{2}}}}\]sq. unit.
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