Key Concepts and Formulas

• Area between curves: If $f(y) \ge g(y)$ on $[c, d]$, the area is $A = \int_c^d (f(y) - g(y)) \, dy$.
• Solving quadratic equations.
• Basic integration rules for polynomial functions.

Step-by-Step Solution

Step 1: Find the points of intersection

We need to find where the parabola $y^2 = 4\lambda x$ and the line $y = \lambda x$ intersect. To do this, substitute $x = \frac{y}{\lambda}$ (from the equation of the line) into the equation of the parabola:

$y^2 = 4\lambda \left(\frac{y}{\lambda}\right)$ $y^2 = 4y$ $y^2 - 4y = 0$ $y(y - 4) = 0$

This gives us $y = 0$ and $y = 4$. So the points of intersection occur at $y=0$ and $y=4$.

Step 2: Express x in terms of y for both equations

From $y^2 = 4\lambda x$, we get $x = \frac{y^2}{4\lambda}$. From $y = \lambda x$, we get $x = \frac{y}{\lambda}$.

Step 3: Set up the integral for the area

Since we have $x$ as a function of $y$, we integrate with respect to $y$. We want to find the area between the curves from $y=0$ to $y=4$. On this interval, the line $x = \frac{y}{\lambda}$ is to the right of the parabola $x = \frac{y^2}{4\lambda}$. Thus, we have:

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

Step 4: Evaluate the integral

$A = \frac{1}{\lambda} \int_0^4 \left(y - \frac{y^2}{4}\right) dy$ $A = \frac{1}{\lambda} \left[\frac{y^2}{2} - \frac{y^3}{12}\right]_0^4$ $A = \frac{1}{\lambda} \left[\frac{4^2}{2} - \frac{4^3}{12} - (0 - 0)\right]$ $A = \frac{1}{\lambda} \left[\frac{16}{2} - \frac{64}{12}\right]$ $A = \frac{1}{\lambda} \left[8 - \frac{16}{3}\right]$ $A = \frac{1}{\lambda} \left[\frac{24 - 16}{3}\right]$ $A = \frac{1}{\lambda} \left[\frac{8}{3}\right]$ $A = \frac{8}{3\lambda}$

Step 5: Solve for λ

We are given that the area is $\frac{1}{9}$. Therefore, $\frac{8}{3\lambda} = \frac{1}{9}$ $3\lambda = 72$ $\lambda = \frac{72}{3}$ $\lambda = 24$

Step 6: Recheck the intersection and area. The intersection points are at $y=0$ and $y=4$. The area is given by $A = \int_0^4 (\frac{y}{\lambda} - \frac{y^2}{4\lambda}) dy = \frac{8}{3\lambda} = \frac{1}{9}$ Solving for $\lambda$, we have $\lambda = \frac{8}{3} \times 9 = 24$.

Step 7: Identify the mistake and correct the calculation

The previous calculation is incorrect. We are given that the area is $\frac{1}{9}$. Therefore, $\frac{8}{3\lambda} = \frac{1}{9}$ $3\lambda = 72$ $\lambda = \frac{72}{3} = 24$

There appears to be no error in the calculation.

Step 8: Correct the mistake in the intersection calculation Let's re-check the intersection calculation. We have $y^2 = 4\lambda x$ and $y = \lambda x$. Substituting $x = y/\lambda$ gives $y^2 = 4\lambda(y/\lambda) = 4y$. So $y^2 - 4y = 0$, so $y(y-4) = 0$. Therefore $y=0$ or $y=4$. The corresponding $x$ values are $x=0$ and $x=4/\lambda$. The area is given by $\int_0^{4/\lambda} (\sqrt{4\lambda x} - \lambda x) dx = \int_0^{4/\lambda} (2\sqrt{\lambda}\sqrt{x} - \lambda x) dx$. Let $y = \lambda x$. Then $x = y/\lambda$. When $x=0$, $y=0$. When $x=4/\lambda$, $y=4$. So we have $\int_0^4 (\frac{y}{\lambda} - \frac{y^2}{4\lambda}) dy = \frac{1}{9}$. The first term should be $\sqrt{4\lambda x}$, not $\frac{y}{\lambda}$.

Step 9: Correct the Integration

The correct integral is: $\int_0^{4/\lambda} (\sqrt{4\lambda x} - \lambda x) \, dx = \frac{1}{9}$ $\int_0^{4/\lambda} (2\sqrt{\lambda} x^{1/2} - \lambda x) \, dx = \frac{1}{9}$ $\left[ 2\sqrt{\lambda} \frac{2}{3} x^{3/2} - \lambda \frac{x^2}{2} \right]_0^{4/\lambda} = \frac{1}{9}$ $\frac{4\sqrt{\lambda}}{3} \left(\frac{4}{\lambda}\right)^{3/2} - \frac{\lambda}{2} \left(\frac{4}{\lambda}\right)^2 = \frac{1}{9}$ $\frac{4\sqrt{\lambda}}{3} \frac{8}{\lambda\sqrt{\lambda}} - \frac{\lambda}{2} \frac{16}{\lambda^2} = \frac{1}{9}$ $\frac{32}{3\lambda} - \frac{8}{\lambda} = \frac{1}{9}$ $\frac{32 - 24}{3\lambda} = \frac{1}{9}$ $\frac{8}{3\lambda} = \frac{1}{9}$ $3\lambda = 72$ $\lambda = 24$

The previous calculation still arrives at 24. The integral is correct.

Step 10: Check again

Let's re-evaluate the area calculation. Area = $\int_0^{4/\lambda} (2\sqrt{\lambda x} - \lambda x)dx = [\frac{4}{3}\sqrt{\lambda} x^{3/2} - \frac{\lambda x^2}{2}]_0^{4/\lambda} = \frac{4}{3}\sqrt{\lambda} (\frac{4}{\lambda})^{3/2} - \frac{\lambda}{2} (\frac{4}{\lambda})^2 = \frac{4}{3}\sqrt{\lambda} \frac{8}{\lambda\sqrt{\lambda}} - \frac{\lambda}{2} \frac{16}{\lambda^2} = \frac{32}{3\lambda} - \frac{8}{\lambda} = \frac{32 - 24}{3\lambda} = \frac{8}{3\lambda}$. So $\frac{8}{3\lambda} = \frac{1}{9}$. $3\lambda = 72$, $\lambda = 24$.

Step 11: Find the Error

The initial question states $y^2=4\lambda x$ and $y = \lambda x$. The area is $\frac{1}{9}$. We calculated the points of intersection correctly as $(0, 0)$ and $(\frac{4}{\lambda}, 4)$. Also, $\int_0^{4/\lambda} (\sqrt{4\lambda x} - \lambda x)dx = \frac{1}{9}$. So $\frac{8}{3\lambda} = \frac{1}{9}$. Thus $\lambda = 24$.

Let $y=4$, $x = \frac{4}{\lambda}$. So $\frac{16}{4\lambda} = \frac{1}{9}$. Then $36 \lambda x - 36 \lambda x = 0$.

The error must be in the question itself. Let's assume the correct answer is $4\sqrt{3}$. Then $\frac{8}{3\lambda} = \frac{1}{9}$, so $\frac{8}{3 * 4\sqrt{3}} = \frac{2}{3\sqrt{3}}$. This is not $\frac{1}{9}$.

Step 12: Retrace the Steps

We are given that the area is $\frac{1}{9}$. $\int_0^{4/\lambda} (2\sqrt{\lambda x} - \lambda x) dx = \frac{1}{9}$ $\left[ \frac{4}{3} \sqrt{\lambda} x^{3/2} - \frac{\lambda x^2}{2} \right]_0^{4/\lambda} = \frac{1}{9}$ $\frac{4}{3} \sqrt{\lambda} \left( \frac{4}{\lambda} \right)^{3/2} - \frac{\lambda}{2} \left( \frac{4}{\lambda} \right)^2 = \frac{1}{9}$ $\frac{4}{3} \sqrt{\lambda} \frac{8}{\lambda \sqrt{\lambda}} - \frac{\lambda}{2} \frac{16}{\lambda^2} = \frac{1}{9}$ $\frac{32}{3\lambda} - \frac{8}{\lambda} = \frac{1}{9}$ $\frac{32 - 24}{3\lambda} = \frac{1}{9}$ $\frac{8}{3\lambda} = \frac{1}{9}$ $72 = 3\lambda$ $\lambda = 24$

Step 13: Correct the Solution

The error lies in assuming that the line is above the parabola in the integration. If we have $y^2 = 4\lambda x$ and $y = \lambda x$, then $x = \frac{y^2}{4\lambda}$ and $x = \frac{y}{\lambda}$. So the area is given by $\int_0^4 (\frac{y}{\lambda} - \frac{y^2}{4\lambda}) dy = \frac{1}{9}$. $\frac{1}{\lambda} [\frac{y^2}{2} - \frac{y^3}{12}]_0^4 = \frac{1}{\lambda} [\frac{16}{2} - \frac{64}{12}] = \frac{1}{\lambda} [8 - \frac{16}{3}] = \frac{1}{\lambda} \frac{8}{3} = \frac{8}{3\lambda} = \frac{1}{9}$. So $\lambda = 24$. This doesn't match the given answer.

Let's assume the given answer is correct: $\lambda = 4\sqrt{3}$. Then the area would be $\frac{8}{3 * 4\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$. This does not equal $\frac{1}{9}$.

Step 14: Find the correct solution by working backward from the answer. The intersection points are at $y=0$ and $y=4$. So $y=4$, $x = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}$. $y^2 = 4\lambda x$. So $16 = 4 * 4\sqrt{3} * \frac{1}{\sqrt{3}}$. So $16 = 16$. The area should be $\frac{1}{9}$. $A = \int_0^4 (\frac{y}{\lambda} - \frac{y^2}{4\lambda}) dy = \frac{8}{3\lambda} = \frac{1}{9}$ $A = \frac{8}{3(4\sqrt{3})} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$ So the area should be $\frac{2\sqrt{3}}{9}$ if $\lambda = 4\sqrt{3}$.

The question must be incorrect since $\lambda=24$ and $\lambda = 4\sqrt{3}$ lead to different areas.

Common Mistakes & Tips

• Always find the points of intersection accurately.
• Carefully determine which function is greater than the other in the interval of integration. A quick sketch can help.
• Double-check your integration and algebraic manipulations.

Summary

We set up and evaluated the integral for the area between the parabola $y^2 = 4\lambda x$ and the line $y = \lambda x$. We equated this area to the given value of $\frac{1}{9}$ and solved for $\lambda$. The calculations lead to $\lambda=24$, however, the given correct answer is $\lambda = 4\sqrt{3}$. There may be an error in the question itself.

Final Answer

The final answer is \boxed{24}, which is not among the options provided. Since the correct answer is given as (A) $4\sqrt{3}$, and the solution yields $\lambda=24$, there seems to be an inconsistency with the question. The derivation is correct, thus the answer is 24.