Equation of given parabola
\(y^2=16 a x\)
and the line \(y=4 m x\)


On simplifying the equation of parabola and line, the \(y\)-coordinate of point \(A\) is \(\frac{4 a}{m}\)
\(\therefore\) The area bounded by given curves
\(\begin{aligned}
& =\int_0^{\frac{4 a}{m}}\left(\frac{y}{4 m}-\frac{y^2}{16 a}\right) d y=\frac{a^2}{12} \quad \text{(given)} \\
& \Rightarrow \quad\left[\frac{y^2}{8 m}-\frac{y^3}{48 a}\right]_0^{\frac{4 a}{m}}=\frac{a^2}{12} \\
& \Rightarrow \quad \frac{16 a^2}{8 m^3}-\frac{64 a^3}{48 m^3 a}=\frac{a^2}{12} \\
& \Rightarrow \quad \frac{2}{m^3}-\frac{4}{3 m^3}=\frac{1}{12} \\
& \Rightarrow \quad \frac{6-4}{3 m^3}=\frac{1}{12} \\
& \Rightarrow \quad m^2=8 \Rightarrow m=2 \\
\end{aligned}\)
Hence, option (d) is correct.