Let the line `y = mx` intersects the curve `y^2 = x` at P and tangent to `y^2 = x` at P intersects x-axis at Q. If area (`triangle`OPQ) = 4, find `m (m gt 0)`.

Answer

To solve the problem, we need to find the value of \( m \) such that the area of triangle \( OPQ \) is equal to 4, where \( O \) is the origin, \( P \) is the intersection of the line \( y = mx \) and the curve \( y^2 = x \), and \( Q \) is the point where the tangent to the curve at point \( P \) intersects the x-axis.

### Step-by-Step Solution:

1. **Find the Intersection Point \( P \)**:
We start with the equations:
\[ y^2 = x \quad \text{(1)} \]
\[ y = mx \quad \text{(2)} \]
Substitute equation (2) into equation (1):
\[ (mx)^2 = x \]
This simplifies to:
\[ m^2 x^2 = x \]
Rearranging gives:
\[ m^2 x^2 - x = 0 \]
Factoring out \( x \):
\[ x(m^2 x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = \frac{1}{m^2} \). Since we are looking for the intersection point other than the origin, we take:
\[ x = \frac{1}{m^2} \]
Now substituting \( x \) back into equation (2) to find \( y \):
\[ y = m \cdot \frac{1}{m^2} = \frac{1}{m} \]
Therefore, the coordinates of point \( P \) are:
\[ P\left(\frac{1}{m^2}, \frac{1}{m}\right) \]

2. **Find the Tangent at Point \( P \)**:
The slope of the tangent to the curve \( y^2 = x \) can be found by differentiating:
\[ 2y \frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{2y} \]
At point \( P \), where \( y = \frac{1}{m} \):
\[ \frac{dy}{dx} = \frac{1}{2 \cdot \frac{1}{m}} = \frac{m}{2} \]
The equation of the tangent line at point \( P \) is given by:
\[ y - \frac{1}{m} = \frac{m}{2}\left(x - \frac{1}{m^2}\right) \]
Rearranging gives:
\[ y = \frac{m}{2}x - \frac{1}{2m} + \frac{1}{m} \]
Simplifying:
\[ y = \frac{m}{2}x + \frac{1}{2m} \]

3. **Find the Point \( Q \)**:
To find where the tangent intersects the x-axis, set \( y = 0 \):
\[ 0 = \frac{m}{2}x + \frac{1}{2m} \]
Solving for \( x \):
\[ \frac{m}{2}x = -\frac{1}{2m} \quad \Rightarrow \quad x = -\frac{1}{m^2} \]
Thus, the coordinates of point \( Q \) are:
\[ Q\left(-\frac{1}{m^2}, 0\right) \]

4. **Calculate the Area of Triangle \( OPQ \)**:
The area \( A \) of triangle \( OPQ \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Here, \( O(0, 0) \), \( P\left(\frac{1}{m^2}, \frac{1}{m}\right) \), and \( Q\left(-\frac{1}{m^2}, 0\right) \):
\[ A = \frac{1}{2} \left| 0\left(\frac{1}{m} - 0\right) + \frac{1}{m^2}(0 - 0) + \left(-\frac{1}{m^2}\right)\left(0 - \frac{1}{m}\right) \right| \]
Simplifying:
\[ A = \frac{1}{2} \left| -\frac{1}{m^2} \cdot -\frac{1}{m} \right| = \frac{1}{2} \cdot \frac{1}{m^3} \]
Setting this equal to 4:
\[ \frac{1}{2m^3} = 4 \quad \Rightarrow \quad 1 = 8m^3 \quad \Rightarrow \quad m^3 = \frac{1}{8} \quad \Rightarrow \quad m = \frac{1}{2} \]

### Final Answer:
Thus, the value of \( m \) is:
\[ \boxed{\frac{1}{2}} \]