If area bounded by the curve `y^(2)=4ax` and `y=mx` is `a^(2)//3` , then the value of m, is

Answer

To solve the problem, we need to find the value of \( m \) such that the area bounded by the curves \( y^2 = 4ax \) (a parabola) and \( y = mx \) (a straight line) is equal to \( \frac{a^2}{3} \).

### Step-by-Step Solution:

1. **Identify the curves**:
- The first curve is \( y^2 = 4ax \), which is a parabola opening to the right.
- The second curve is \( y = mx \), which is a straight line passing through the origin.

2. **Find the points of intersection**:
- To find the points of intersection, substitute \( y = mx \) into \( y^2 = 4ax \):
\[ (mx)^2 = 4ax \]
\[ m^2x^2 - 4ax = 0 \]
\[ x(m^2x - 4a) = 0 \]
- This gives us \( x = 0 \) (the origin) and \( x = \frac{4a}{m^2} \).

3. **Find the corresponding y-coordinate**:
- For \( x = \frac{4a}{m^2} \):
\[ y = m \left(\frac{4a}{m^2}\right) = \frac{4a}{m} \]
- Therefore, the points of intersection are \( (0, 0) \) and \( \left(\frac{4a}{m^2}, \frac{4a}{m}\right) \).

4. **Set up the area integral**:
- The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{4a}{m^2} \) is given by:
\[ A = \int_0^{\frac{4a}{m^2}} \left(\sqrt{4ax} - mx\right) \, dx \]
- Here, \( \sqrt{4ax} \) is the upper curve (the parabola) and \( mx \) is the lower curve (the line).

5. **Calculate the integral**:
- The integral can be computed as follows:
\[ A = \int_0^{\frac{4a}{m^2}} \left(2\sqrt{ax} - mx\right) \, dx \]
- The integral of \( 2\sqrt{ax} \) is:
\[ \int 2\sqrt{ax} \, dx = \frac{4}{3} a^{1/2} x^{3/2} \]
- The integral of \( mx \) is:
\[ \int mx \, dx = \frac{mx^2}{2} \]

6. **Evaluate the definite integral**:
- Evaluating from \( 0 \) to \( \frac{4a}{m^2} \):
\[ A = \left[\frac{4}{3} a^{1/2} \left(\frac{4a}{m^2}\right)^{3/2} - \frac{m}{2} \left(\frac{4a}{m^2}\right)^2\right] \]
- Simplifying this gives:
\[ A = \frac{4}{3} a^{1/2} \cdot \frac{8a^{3/2}}{m^3} - \frac{8a^2}{m^2} \]
\[ A = \frac{32a^2}{3m^3} - \frac{8a^2}{m^2} \]

7. **Set the area equal to \( \frac{a^2}{3} \)**:
- We set the area equal to \( \frac{a^2}{3} \):
\[ \frac{32a^2}{3m^3} - \frac{8a^2}{m^2} = \frac{a^2}{3} \]
- Dividing through by \( a^2 \) (assuming \( a \neq 0 \)):
\[ \frac{32}{3m^3} - \frac{8}{m^2} = \frac{1}{3} \]

8. **Multiply through by \( 3m^3 \)** to eliminate the fractions:
- This gives:
\[ 32 - 24m = m^3 \]
- Rearranging gives:
\[ m^3 + 24m - 32 = 0 \]

9. **Solve the cubic equation**:
- By trial and error or using the Rational Root Theorem, we find \( m = 2 \) is a root.
- Factoring out \( m - 2 \) gives:
\[ (m - 2)(m^2 + 2m + 16) = 0 \]
- The quadratic \( m^2 + 2m + 16 \) has no real roots (discriminant is negative).

10. **Conclusion**:
- The only real solution is \( m = 2 \).

### Final Answer:
The value of \( m \) is \( 2 \).