Let a line $$y = mx$$ $$(m \gt 0)$$, intersect the parabola, $$y^2 = x$$, at a point P, other than the origin. Let the tangent to it at P, meet the x-axis at the point Q. If area ($$\triangle OPQ$$) = 4 square unit, then m is equal to

We have a straight line $$y = mx$$ with the given condition $$m \gt 0$$. This line meets the parabola $$y^2 = x$$ at two points. One of them is obviously the origin $$O(0,0)$$. To obtain the other point, we substitute $$x = \dfrac{y}{m}$$ (from $$y = mx$$) into the parabola’s equation:

$$y^2 \;=\; x \;=\; \dfrac{y}{m}.$$

Re-arranging,

$$y^2 - \dfrac{y}{m} = 0 \;\;\Longrightarrow\;\; y\Bigl(y - \dfrac1m\Bigr) = 0.$$

This gives two possible values of $$y$$:

$$y = 0 \quad\text{or}\quad y = \dfrac1m.$$

The value $$y = 0$$ corresponds to the origin, so the non-trivial point of intersection is obtained from

$$y_P = \dfrac1m.$$

For this $$y_P$$, the corresponding $$x_P$$ is found from $$x = \dfrac{y}{m}$$:

$$x_P = \dfrac{1/m}{m} = \dfrac1{m^2}.$$

Thus, the coordinates of the required point $$P$$ are

$$P\!\Bigl(\dfrac1{m^2},\; \dfrac1m\Bigr).$$

Next we need the tangent to the parabola $$y^2 = x$$ at the point $$P$$. First we find the slope of the tangent. Differentiating $$y^2 = x$$ implicitly with respect to $$x$$:

$$2y\dfrac{dy}{dx} = 1 \;\;\Longrightarrow\;\; \dfrac{dy}{dx} = \dfrac1{2y}.$$

At $$P$$, where $$y = \dfrac1m$$, the slope of the tangent is

$$m_{\text{tangent}} = \dfrac1{2\bigl(1/m\bigr)} = \dfrac1{2/m} = \dfrac{m}{2}.$$

The equation of the tangent at $$P\bigl(x_P,y_P\bigr)$$ is, therefore,

$$y - y_P = m_{\text{tangent}}\bigl(x - x_P\bigr),$$

that is,

$$y - \dfrac1m = \dfrac{m}{2}\!\left( x - \dfrac1{m^2} \right).$$

This tangent meets the $$x$$-axis where $$y = 0$$. Putting $$y = 0$$ in the above equation gives

$$0 - \dfrac1m = \dfrac{m}{2}\!\left( x_Q - \dfrac1{m^2} \right).$$

We solve for $$x_Q$$ step by step:

$$-\dfrac1m = \dfrac{m}{2}\,x_Q - \dfrac{m}{2}\cdot\dfrac1{m^2}$$ $$-\dfrac1m = \dfrac{m}{2}\,x_Q - \dfrac1{2m}$$

Now bring the second term on the right to the left:

$$-\dfrac1m + \dfrac1{2m} = \dfrac{m}{2}\,x_Q$$ $$-\dfrac1{2m} = \dfrac{m}{2}\,x_Q$$

Multiplying both sides by $$\dfrac2m$$:

$$x_Q = -\,\dfrac2{m^2}\cdot\dfrac12 = -\dfrac1{m^2}.$$

Thus, the tangent meets the $$x$$-axis at

$$Q\!\Bigl(-\dfrac1{m^2},\,0\Bigr).$$

Now we have the three vertices of $$\triangle OPQ$$:

$$O(0,0), \quad P\!\Bigl(\dfrac1{m^2},\,\dfrac1m\Bigr), \quad Q\!\Bigl(-\dfrac1{m^2},\,0\Bigr).$$

The area of a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ is given by the determinant formula

$$\text{Area} \;=\; \dfrac12\Bigl|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\Bigr|.$$

Substituting $$(x_1,y_1) = (0,0)$$, $$(x_2,y_2) = \bigl(\tfrac1{m^2}, \tfrac1m\bigr)$$ and $$(x_3,y_3) = \bigl(-\tfrac1{m^2},0\bigr)$$, we get

$$\text{Area} = \dfrac12\Bigl|\; 0\bigl(\tfrac1m - 0\bigr) \;+\; \dfrac1{m^2}\bigl(0 - 0\bigr) \;+\; \Bigl(-\dfrac1{m^2}\Bigr)\bigl(0 - \tfrac1m\bigr) \Bigr|.$$

The first two terms vanish, so

$$\text{Area} = \dfrac12\Bigl|\, \Bigl(-\dfrac1{m^2}\Bigr)\Bigl(-\dfrac1m\Bigr) \Bigr| = \dfrac12\Bigl(\dfrac1{m^3}\Bigr) = \dfrac1{2m^3}.$$

According to the question, this area equals $$4$$ square units:

$$\dfrac1{2m^3} = 4.$$

Multiplying both sides by $$2m^3$$:

$$1 = 8m^3 \;\;\Longrightarrow\;\; m^3 = \dfrac18.$$

Taking the positive cube root (because $$m \gt 0$$), we obtain

$$m = \dfrac12 = 0.5.$$

So, the answer is $$m = 0.5.$$