If the area (in sq. units) bounded by the parabola y^2=4λ x and the line y=λ x, λ >0, is 1 9, then λ is equal to

The given equations are ${y}^{2}=4\lambda x$y2=4λx and $y=\lambda x$y=λx

Using above equations, we can write

$${(\lambda x)}^{2}=4\lambda x\Rightarrow {\lambda }^{2}{x}^{2}-4\lambda x=0$$(λx)2=4λx⇒λ2x2−4λx=0

$$\Rightarrow \lambda x(\lambda x-4)=0$$⇒λx(λx−4)=0

\Rightarrow x=0&x=\frac{4}{\lambda }

Area $\displaystyle =\frac{1}{9}=\int _{0}^{\frac{4}{\lambda }}(\sqrt{4\lambda x}-\lambda x)dx$=91​=∫0λ4​​(4λx​−λx)dx

$$\Rightarrow {(\frac{\sqrt{4\lambda }{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\lambda {x}^{2}}{2})}_{0}^{\frac{4}{\lambda }}=\frac{1}{9}$$⇒(23​4λ​x23​​−2λx2​)0λ4​​=91​

$$\Rightarrow \lambda =24$$⇒λ=24

Hence, the required value is $24$24.