Given curves

The parabola is $y^2=4\lambda x$ and the line is $y=\lambda x, \quad \lambda>0.$

Write both as $x$ in terms of $y$

From the parabola: $x=\frac{y^2}{4\lambda}$

From the line: $y=\lambda x \implies x=\frac{y}{\lambda}$

Find points of intersection

Substitute $x=\frac{y}{\lambda}$ into $y^2=4\lambda x$: $y^2=4\lambda\left(\frac{y}{\lambda}\right)=4y$

So, $y^2-4y=0$ $y(y-4)=0$

Hence, $y=0 \quad \text{or} \quad y=4.$

Corresponding $x$ values from $x=\frac{y}{\lambda}$ are:

• for $y=0$, $x=0$
• for $y=4$, $x=\frac{4}{\lambda}$

So the curves intersect at $(0,0) \quad \text{and} \quad \left(\frac{4}{\lambda},4\right).$

Determine which curve is to the right

For $0<y<4$: $x_{\text{line}}=\frac{y}{\lambda}, \qquad x_{\text{parabola}}=\frac{y^2}{4\lambda}$

Since for $0<y<4$, $\frac{y}{\lambda}>\frac{y^2}{4\lambda},$ the line lies to the right of the parabola.

Area between the curves

Therefore, $A=\int_0^4 \left(\frac{y}{\lambda}-\frac{y^2}{4\lambda}\right)dy$

Factor out $\frac{1}{\lambda}$: $A=\frac{1}{\lambda}\int_0^4\left(y-\frac{y^2}{4}\right)dy$

Now integrate: $\int\left(y-\frac{y^2}{4}\right)dy=\frac{y^2}{2}-\frac{y^3}{12}$

So, $A=\frac{1}{\lambda}\left[\frac{y^2}{2}-\frac{y^3}{12}\right]_0^4$

$A=\frac{1}{\lambda}\left(\frac{16}{2}-\frac{64}{12}\right)$

$A=\frac{1}{\lambda}\left(8-\frac{16}{3}\right)$

$A=\frac{1}{\lambda}\cdot\frac{8}{3}$

Hence, $A=\frac{8}{3\lambda}$

Use the given area

Given that the area is $\frac{1}{9}$, $\frac{8}{3\lambda}=\frac{1}{9}$

Cross-multiplying: $72=3\lambda$ $\lambda=24$

Check options

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

So the correct option is D.