question_answer

If \[{{A}_{1}}\]  is the area of the parabola \[{{y}^{2}}=4ax\] lying between vertex and the latusrectum and \[{{A}_{2}}\] is the area between the latusrectum and the double ordinate x = 2a, then \[{{A}_{1}}/{{A}_{2}}\]is equal to

A)  \[(2\sqrt{2}-1)\]                             

B)  \[\frac{1}{7}(2\sqrt{2}+1)\]

C)  \[\frac{1}{7}(2\sqrt{2}-1)\]                        

D)  None of these

Correct Answer: B

Solution :

\[{{A}_{1}}=2\int_{0}^{a}{\sqrt{4ax}}dx=4\sqrt{a}\int_{0}^{a}{\sqrt{x}}dx\] \[=4\sqrt{a}\frac{2}{3}[{{x}^{3/2}}]_{0}^{a}\]                     \[=\frac{8{{a}^{2}}}{3}\text{sq}\,\text{unit}\] and\[{{A}_{2}}=2\left[ \int_{a}^{2a}{\sqrt{4ax}}dx \right]\]                  \[=\frac{8\sqrt{a}}{3}[{{x}^{3/2}}]_{a}^{2a}\] \[=\frac{8\sqrt{a}}{3}[2\sqrt{2}-1]\]sq unit \[\therefore \]\[\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{1}{2\sqrt{2}-1}=\frac{2\sqrt{2}+1}{7}\]