A) 8 sq. units
B) 4 sq. units
C) 4/3 sq. units
D) 8/3 sq. units
Correct Answer: D
Solution :
[d] Point of intersection of \[{{y}^{2}}=4ax\] |
and \[y=ax\] are (0, 0) and \[\left( \frac{4}{a},4 \right)\] |
Given \[\int\limits_{0}^{4}{\left[ \frac{y}{a}-\frac{{{y}^{2}}}{4a} \right]dy=\frac{1}{3}}\] |
\[\Rightarrow \frac{8}{a}-\frac{1}{12a}\times 64=\frac{1}{3}\Rightarrow \frac{8}{3a}=\frac{1}{3}\Rightarrow a=8\] |
So, the parabola is \[{{y}^{2}}=32x\] |
Area enclosed by \[y=4x\] is |
\[\int\limits_{0}^{8}{\left[ \frac{y}{4}-\frac{{{y}^{2}}}{32} \right]dy=\left[ \frac{{{y}^{2}}}{8}-\frac{{{y}^{3}}}{96} \right]}_{0}^{8}=\frac{8}{3}\] |
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