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

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the area bounded by the parabola \( y^2 = 4\lambda x \) and the line \( y = \lambda x \) is \( \frac{1}{9} \).

### Step-by-Step Solution:

1. **Identify the equations**:
- The equation of the parabola is \( y^2 = 4\lambda x \).
- The equation of the line is \( y = \lambda x \).

2. **Find the points of intersection**:
- Set \( y = \lambda x \) into the parabola's equation:
\[ (\lambda x)^2 = 4\lambda x \]
This simplifies to:
\[ \lambda^2 x^2 - 4\lambda x = 0 \]
- Factor out \( \lambda x \):
\[ \lambda x (\lambda x - 4) = 0 \]
- The solutions are:
\[ x = 0 \quad \text{or} \quad x = \frac{4}{\lambda} \]

3. **Set up the area integral**:
- The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{4}{\lambda} \) is given by:
\[ A = \int_0^{\frac{4}{\lambda}} \left( \sqrt{4\lambda x} - \lambda x \right) dx \]

4. **Simplify the integral**:
- The integral becomes:
\[ A = \int_0^{\frac{4}{\lambda}} \left( 2\sqrt{\lambda x} - \lambda x \right) dx \]

5. **Calculate the integral**:
- Split the integral:
\[ A = \int_0^{\frac{4}{\lambda}} 2\sqrt{\lambda x} \, dx - \int_0^{\frac{4}{\lambda}} \lambda x \, dx \]
- For the first integral:
\[ \int 2\sqrt{\lambda x} \, dx = 2\sqrt{\lambda} \cdot \frac{2}{3} x^{3/2} = \frac{4\sqrt{\lambda}}{3} x^{3/2} \]
Evaluating from \( 0 \) to \( \frac{4}{\lambda} \):
\[ = \frac{4\sqrt{\lambda}}{3} \left( \frac{4}{\lambda} \right)^{3/2} = \frac{4\sqrt{\lambda}}{3} \cdot \frac{8}{\lambda^{3/2}} = \frac{32}{3\sqrt{\lambda}} \]
- For the second integral:
\[ \int \lambda x \, dx = \frac{\lambda}{2} x^2 \]
Evaluating from \( 0 \) to \( \frac{4}{\lambda} \):
\[ = \frac{\lambda}{2} \left( \frac{4}{\lambda} \right)^2 = \frac{\lambda}{2} \cdot \frac{16}{\lambda^2} = \frac{8}{\lambda} \]

6. **Combine the results**:
- The area \( A \) is:
\[ A = \frac{32}{3\sqrt{\lambda}} - \frac{8}{\lambda} \]

7. **Set the area equal to \( \frac{1}{9} \)**:
\[ \frac{32}{3\sqrt{\lambda}} - \frac{8}{\lambda} = \frac{1}{9} \]

8. **Clear the fractions**:
- Multiply through by \( 9\lambda \):
\[ 96\lambda - 72\sqrt{\lambda} = \lambda \]
- Rearranging gives:
\[ 95\lambda - 72\sqrt{\lambda} = 0 \]

9. **Factor out \( \sqrt{\lambda} \)**:
\[ \sqrt{\lambda}(95\sqrt{\lambda} - 72) = 0 \]
- Thus, \( \sqrt{\lambda} = \frac{72}{95} \) or \( \lambda = \left( \frac{72}{95} \right)^2 \).

10. **Calculate \( \lambda \)**:
\[ \lambda = \frac{5184}{9025} \]

### Final Answer:
\[ \lambda = \frac{5184}{9025} \]