If the area bounded by the curve y^2=16x and the line y = mx is 2/3 s…

The correct answer is Option (4) → 4

Given equations are,

$y^2=16x$   ...(1)

$y = mx$   ...(2)

for $x≠0$,

$(mx)^2=16x$

$⇒m^2x=16⇒x=\frac{16}{m^2}$

and,

$y=m×\frac{16}{m^2}=\frac{16}{m}$

Thus the non-zero intersection point is $\left(\frac{16}{m^2},\frac{16}{m}\right)$

$∴A=\int\limits_0^{16/m^2}(4\sqrt{x}-mx)dx$

$=\left[\frac{8}{3}x^{3/2}-m\frac{x^2}{2}\right]_0^{16/m^2}$

$=\frac{512}{3m^3}-\frac{128}{m^3}=\frac{128}{3m^3}$

$⇒\frac{128}{3m^3}=\frac{2}{3}$

$⇒m^3=\frac{128}{2}=64$

$⇒m=4$