question_answer

If the area enclosed by \[{{y}^{2}}=4ax\] and line \[y=ax\]is 1/3 sq. units , then the area enclosed by \[y=4x\]with same parabola is

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

\[\int\limits_{0}^{8}{\left[ \frac{y}{4}-\frac{{{y}^{2}}}{32} \right]dy=\left[ \frac{{{y}^{2}}}{8}-\frac{{{y}^{3}}}{96} \right]}_{0}^{8}=\frac{8}{3}\]