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

Answer 1

The correct option is C $24$

Given parabola $y^2=4\lambda x$ and the line $y=\lambda x$
Putting $y=\lambda x$ in $y^2=4\lambda x,$ we get $x=0,\frac{4}{\lambda }$
$\therefore$ Required area $=\underset{0}{\overset{\frac{4}{\lambda }}{\int }}(2\sqrt{\lambda x}-\lambda x)dx$
$={[\frac{2\sqrt{\lambda }\cdot x^{3/2}}{3/2}-\frac{\lambda \cdot x^2}{2}]}_0^{4/\lambda }$
$=\frac{32}{3\lambda }-\frac{8}{\lambda }$
$=\frac{8}{3\lambda }=\frac{1}{9}\text{(Given)}\Rightarrow \lambda =24$