question_answer

Area included   between the two curves, \[{{y}^{2}}=4ax\]and \[{{x}^{2}}-4ay\]is equal to:

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\[=\int_{a}^{4a}{({{y}_{2}}-{{y}_{1}})}\,dx\] \[=\int_{0}^{4a}{\left( \sqrt{4ax}-\frac{{{x}^{2}}}{4a} \right)}dx\] \[=\int_{0}^{4a}{2{{a}^{1/2}}{{x}^{1/2}}dx-\int_{0}^{4a}{\frac{{{x}^{2}}}{4a}dx}}\] \[=2{{a}^{1/2}}\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{4a}-\left[ \frac{{{x}^{3}}}{12a} \right]_{0}^{4a}\] \[=\frac{4}{3}{{a}^{1/2}}[{{(4a)}^{3/2}}-0]-\frac{1}{12a}[{{(4a)}^{3}}]\]                 \[=\frac{4}{3}{{a}^{2}}(B)-\frac{1}{12a}(64{{a}^{3}})\] \[=\frac{32{{a}^{2}}}{3}-\frac{16}{3}{{a}^{2}}\] \[=\frac{16}{3}{{a}^{2}}\,\text{sq}\,\text{unit}\]