Question

If the area (in sq. units) bounded by the parabola $y^2=4\lambda x$ and the line $y=\lambda x,\lambda >0$, is $\frac{1}{9}$ , then $\lambda$ is equal to :

• A. $4\sqrt{3}$
• B. $2\sqrt{6}$
• C. $24$
• D. $48$

Answer 1

The correct option is C $24$

$y^2=4\lambda x$, $y=\lambda x$
Point of intersection are
$x=0,x=\frac{4}{\lambda }$


Required area $=\underset{0}{\overset{4/\lambda }{\int }}(\sqrt{4\lambda x}-\lambda x)dx=\frac{1}{9}$
$\Rightarrow 2\sqrt{\lambda }\times {\left[\frac{2x^{3/2}}{3}\right]}_0^{4/\lambda }-\lambda {\left[\frac{x^2}{2}\right]}_0^{4/\lambda }=\frac{1}{9}\Rightarrow \frac{32}{3\lambda }-\frac{8}{\lambda }=\frac{1}{9}\Rightarrow \lambda =24$