A) \[\frac{16}{3}{{a}^{2}}\text{sq units}\]
B) \[\frac{8}{3}\text{sq}\,\text{units}\]
C) \[\frac{\text{4}}{\text{3}}{{\text{a}}^{\text{2}}}\text{sq}\,\text{units}\]
D) None of these
Correct Answer: A
Solution :
The focus of the parabolas \[{{y}^{2}}=4a(x+a)\]and\[{{y}^{2}}=-4a(x-a)\] are \[(-a,0)\] and \[(a,0)\] respectively. The required area is shown in the figure which is given by as following. Area of \[ABCD=2.\]area of \[AOCB\] \[=4.\]area of AOB \[=4.\int_{-a}^{0}{\sqrt{4a(x+a)}}dx\] \[=4.2\sqrt{a}\int_{-a}^{0}{\sqrt{x+a}}\,dx\] \[=8\sqrt{a}\left[ \frac{2}{3}{{(x+a)}^{3/2}} \right]_{-a}^{0}\] \[\Rightarrow \]Area of \[ABCD=8\sqrt{a}.\frac{2}{3}({{a}^{3/2}}-0)\] \[=\frac{16}{3}\sqrt{a}.{{a}^{3/2}}\] \[=\frac{16}{3}{{a}^{2}}\] which is the required area.You need to login to perform this action.
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