If the area bounded by the parabola $y^2=16 a x$ and the line $y=4 m x$ is $\frac{a^2}{12}$ sq. units, then the value of ' $m$ ' is

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Aicharya · Accepted Answer

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.

If the area bounded by the parabola \(y^2=16 a x\) and the line \(y=4 m x\) is \(\frac{a^2}{12}\) sq. units,…

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