A) \[\frac{32}{3}{{a}^{2}}\text{sq}\,\text{unit}\]
B) \[\frac{64}{9}\,\text{sq}\,\text{unit}\]
C) \[\frac{32}{3}\text{sq}\,\text{unit}\]
D) \[\frac{16}{3}{{a}^{2}}\text{sq}\,\text{unit}\]
Correct Answer: D
Solution :
Given two equations of curves are \[{{y}^{2}}=4a\,x\]and \[{{x}^{2}}=4ay.\] The point of intersection of two curves are \[(0,0)\]and\[(4a,4a).\] \[\therefore \]Required area\[=\int_{a}^{4a}{({{y}_{2}}-{{y}_{1}})}\,dx\] \[=\int_{0}^{4a}{\left( \sqrt{4ax}-\frac{{{x}^{2}}}{4a} \right)}dx\] \[=\int_{0}^{4a}{2{{a}^{1/2}}{{x}^{1/2}}dx-\int_{0}^{4a}{\frac{{{x}^{2}}}{4a}dx}}\] \[=2{{a}^{1/2}}\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{4a}-\left[ \frac{{{x}^{3}}}{12a} \right]_{0}^{4a}\] \[=\frac{4}{3}{{a}^{1/2}}[{{(4a)}^{3/2}}-0]-\frac{1}{12a}[{{(4a)}^{3}}]\] \[=\frac{4}{3}{{a}^{2}}(B)-\frac{1}{12a}(64{{a}^{3}})\] \[=\frac{32{{a}^{2}}}{3}-\frac{16}{3}{{a}^{2}}\] \[=\frac{16}{3}{{a}^{2}}\,\text{sq}\,\text{unit}\]You need to login to perform this action.
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