question_answer

The area bounded by the parabola \[{{y}^{2}}=4ax\] and \[{{\mathbf{x}}^{\mathbf{2}}}=\mathbf{4ay}\] is

A)  \[\frac{8{{a}^{2}}}{3}\]                    

B)  \[\frac{16{{a}^{2}}}{3}\]          

C)  \[8{{a}^{3}}\]                                   

D)  None of these

Correct Answer: B

Solution :

[b] Given equation of parabola be \[{{y}^{2}}=4ax\]                               .............(1) \[{{x}^{2}}=4ay\]                         .............(2) Solving (1) & (2), we have \[{{y}^{2}}=4a.\sqrt{4ay}=8a.\sqrt{ay}\] \[\Rightarrow {{y}^{4}}=64{{a}^{3}}y\] \[\Rightarrow y\left( {{y}^{3}}-64{{a}^{3}} \right)=0\] \[\therefore y=0,/\text{ }y=4a\] Putting these values in (2), we have \[\therefore x=0,x=4a.\] \[\therefore \]Intersection points of curves be \[\left( 0,0 \right)\] and \[\left( 4a,4a \right)\] \[\therefore \]Shaded area of the curves \[=\int\limits_{0}^{4a}{({{y}_{1}}-{{y}_{2}})}.dx=\int\limits_{0}^{4a}{\left( \sqrt{4ax}-\frac{{{x}^{2}}}{4a} \right)}.dx\] \[=2.\sqrt{a}\left( \frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}} \right)-\frac{1}{4a}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4a}\] \[=2\sqrt{a}.\frac{2}{3}{{(4a)}^{3/2}}-\frac{1}{4a}.\frac{1}{3}.{{(4a)}^{3}}\] \[=\frac{4\sqrt{a}}{3}.8.{{a}^{3/2}}-\frac{16{{a}^{2}}}{3}\] \[=\frac{32{{a}^{2}}-16{{a}^{2}}}{3}=\frac{16{{a}^{2}}}{3}\] Hence, option [b] is correct.