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

We have the parabola $$y^{2}=4\lambda x$$ and the straight line $$y=\lambda x$$ with $$\lambda>0$$. To locate their common points, we substitute the expression for $$x$$ from the line into the parabola.

From $$y=\lambda x$$ we obtain $$x=\dfrac{y}{\lambda}$$. Putting this in $$y^{2}=4\lambda x$$ gives

$$y^{2}=4\lambda\left(\dfrac{y}{\lambda}\right)=4y.$$

Hence $$y^{2}-4y=0\;\Longrightarrow\;y(y-4)=0,$$ so $$y=0\quad\text{or}\quad y=4.$$

For $$y=0$$ we have $$x=0$$, and for $$y=4$$ we have $$x=\dfrac{4}{\lambda}.$$ Thus the two curves meet at $$\bigl(0,0\bigr)$$ and $$\left(\dfrac{4}{\lambda},4\right).$$

Between these points we must decide which curve lies to the right (larger $$x$$ value). For a representative ordinate, say $$y=2$$, we find $$x_{\text{line}}=\dfrac{2}{\lambda},\qquad x_{\text{parabola}}=\dfrac{2^{2}}{4\lambda}=\dfrac{1}{\lambda},$$ so the line is to the right and the parabola to the left throughout $$0\le y\le4$$.

The formula for the area between two curves expressed as $$x=f_{1}(y)$$ (right) and $$x=f_{2}(y)$$ (left) is $$\text{Area}=\displaystyle\int_{y_1}^{y_2}\bigl[f_{1}(y)-f_{2}(y)\bigr]\;dy.$$

Here $$f_{1}(y)=\dfrac{y}{\lambda},\qquad f_{2}(y)=\dfrac{y^{2}}{4\lambda},\qquad y_{1}=0,\; y_{2}=4.$$ Therefore

$$\text{Area}=\int_{0}^{4}\left[\frac{y}{\lambda}-\frac{y^{2}}{4\lambda}\right]dy =\frac{1}{\lambda}\int_{0}^{4}\left[y-\frac{y^{2}}{4}\right]dy.$$

Combining terms inside the integral,

$$y-\frac{y^{2}}{4}=\frac{4y-y^{2}}{4},$$ so

$$\text{Area}=\frac{1}{4\lambda}\int_{0}^{4}(4y-y^{2})\,dy.$$

Now we evaluate the integral step by step:

$$\int 4y\,dy = 2y^{2},\qquad \int y^{2}\,dy = \frac{y^{3}}{3}.$$ Hence

$$\int_{0}^{4}(4y-y^{2})dy =\left[2y^{2}-\frac{y^{3}}{3}\right]_{0}^{4} =\left(2\cdot4^{2}-\frac{4^{3}}{3}\right)-0 =\left(2\cdot16-\frac{64}{3}\right) =32-\frac{64}{3} =\frac{96-64}{3} =\frac{32}{3}.$$

Substituting this back,

$$\text{Area}=\frac{1}{4\lambda}\times\frac{32}{3}=\frac{32}{12\lambda}=\frac{8}{3\lambda}.$$

The problem tells us that this area equals $$\dfrac{1}{9}$$, so we set

$$\frac{8}{3\lambda}=\frac{1}{9}.$$

Cross-multiplying, $$8\cdot9=3\lambda\cdot1\;\Longrightarrow\;72=3\lambda\;\Longrightarrow\;\lambda=24.$$

Hence, the correct answer is Option 4.