Question

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

Answer 1

Answer: (D)

Point of intersection of $y^2=4ax$
and $y=ax$ are (0,0) and $\left(\frac{4}{a},4\right)$
Given ${\int }_0^4\left[\frac{y}{a}-\frac{y^2}{4a}\right]dy=\frac{1}{3}$
$\Rightarrow \frac{8}{a}-\frac{1}{12a}\times 64=\frac{1}{3}\Rightarrow \frac{8}{3a}=\frac{1}{3}\Rightarrow a=8$
So, the parabola is $y^2=32x$
Area enclosed by y = 4x is
${\int }_0^8\left[\frac{y}{4}-\frac{y^2}{32}\right]dy={\left[\frac{y^2}{8}-\frac{y^3}{96}\right]}_0^8=\frac{8}{3}$