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)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the intersection point P
The line \( y = mx \) intersects the curve \( y^2 = x \). To find the intersection point, we substitute \( y = mx \) into the equation of the curve:

\[ (mx)^2 = x \]

This simplifies to:

\[ m^2 x^2 = x \]

Rearranging gives:

\[ m^2 x^2 - x = 0 \]

Factoring out \( x \):

\[ x(m^2 x - 1) = 0 \]

This gives us two solutions: \( x = 0 \) or \( m^2 x - 1 = 0 \). For \( x = 0 \), \( y = 0 \) (the origin O). The other solution is:

\[ m^2 x = 1 \implies x = \frac{1}{m^2} \]

Substituting back to find \( y \):

\[ y = m \left(\frac{1}{m^2}\right) = \frac{1}{m} \]

Thus, the point \( P \) is:

\[ P\left(\frac{1}{m^2}, \frac{1}{m}\right) \]

### Step 2: Find the equation of the tangent at point P
The equation of the tangent to the curve \( y^2 = x \) at point \( P \) can be derived using the formula for the tangent to the parabola \( y^2 = 4ax \). Here, \( a = \frac{1}{4} \).

The tangent line at point \( (x_0, y_0) \) is given by:

\[ yy_0 = 2a(x + x_0) \]

Substituting \( a = \frac{1}{4} \), \( x_0 = \frac{1}{m^2} \), and \( y_0 = \frac{1}{m} \):

\[ y \cdot \frac{1}{m} = \frac{1}{2}\left(\frac{1}{4}\right)\left(x + \frac{1}{m^2}\right) \]

This simplifies to:

\[ y = \frac{1}{m}\left(\frac{1}{2}\left(\frac{1}{4}\right)\left(x + \frac{1}{m^2}\right)\right) \]

### Step 3: Find the intersection point Q with the x-axis
To find point \( Q \), we set \( y = 0 \) in the tangent equation:

\[ 0 = \frac{1}{m}\left(\frac{1}{2}\left(\frac{1}{4}\right)\left(x + \frac{1}{m^2}\right)\right) \]

Solving for \( x \):

\[ 0 = \frac{1}{2}\left(\frac{1}{4}\right)\left(x + \frac{1}{m^2}\right) \implies x + \frac{1}{m^2} = 0 \implies x = -\frac{1}{m^2} \]

Thus, point \( Q \) is:

\[ Q\left(-\frac{1}{m^2}, 0\right) \]

### Step 4: Calculate the area of triangle OPQ
The area of triangle \( OPQ \) can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is the distance from \( O \) to \( Q \) which is:

\[ \left|-\frac{1}{m^2} - 0\right| = \frac{1}{m^2} \]

The height is the y-coordinate of point \( P \):

\[ \frac{1}{m} \]

Thus, the area is:

\[ \text{Area} = \frac{1}{2} \times \frac{1}{m^2} \times \frac{1}{m} = \frac{1}{2m^3} \]

### Step 5: Set the area equal to 4
Given that the area of triangle \( OPQ \) is 4, we set up the equation:

\[ \frac{1}{2m^3} = 4 \]

Multiplying both sides by \( 2m^3 \):

\[ 1 = 8m^3 \implies m^3 = \frac{1}{8} \implies m = \frac{1}{2} \]

### Final Answer
Thus, the value of \( m \) is:

\[ \boxed{\frac{1}{2}} \]