If the tangent to the curve `y = 1 – x^(2) at x = alpha`, where` 0 lt alpha lt 1,` meets the axes at P and Q. Also `alpha` varies, the minimum value of the area of the triangle OPQ is k times area bounded by the axes and the part of the curve for which `0 lt x lt 1`, then k is equal to

Explanation

To solve the problem, we will follow these steps:

### Step 1: Find the equation of the tangent to the curve at \( x = \alpha \)

The curve is given by:
\[ y = 1 - x^2 \]
At \( x = \alpha \), the corresponding \( y \) value is:
\[ y = 1 - \alpha^2 \]

Next, we need to find the slope of the tangent line at this point. The derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -2x \]
At \( x = \alpha \), the slope \( m \) is:
\[ m = -2\alpha \]

Using the point-slope form of the equation of a line, the equation of the tangent line at \( (\alpha, 1 - \alpha^2) \) is:
\[ y - (1 - \alpha^2) = -2\alpha(x - \alpha) \]
Simplifying this, we get:
\[ y = -2\alpha x + 2\alpha^2 + 1 - \alpha^2 \]
\[ y = -2\alpha x + 1 + \alpha^2 \]

### Step 2: Find the points of intersection with the axes

**Finding point P (x-intercept):**
Set \( y = 0 \):
\[ 0 = -2\alpha x + 1 + \alpha^2 \]
Solving for \( x \):
\[ 2\alpha x = 1 + \alpha^2 \implies x = \frac{1 + \alpha^2}{2\alpha} \]
Thus, point \( P \) is:
\[ P\left(\frac{1 + \alpha^2}{2\alpha}, 0\right) \]

**Finding point Q (y-intercept):**
Set \( x = 0 \):
\[ y = 1 + \alpha^2 \]
Thus, point \( Q \) is:
\[ Q(0, 1 + \alpha^2) \]

### Step 3: Calculate the area of triangle OPQ

The area \( A \) of triangle \( OPQ \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the x-coordinate of point \( P \) and the height is the y-coordinate of point \( Q \):
\[ A = \frac{1}{2} \times \frac{1 + \alpha^2}{2\alpha} \times (1 + \alpha^2) \]
\[ A = \frac{(1 + \alpha^2)^2}{4\alpha} \]

### Step 4: Find the area bounded by the axes and the curve from \( x = 0 \) to \( x = 1 \)

The area \( A_{curve} \) under the curve from \( x = 0 \) to \( x = 1 \) is given by:
\[ A_{curve} = \int_0^1 (1 - x^2) \, dx \]
Calculating this integral:
\[ A_{curve} = \left[ x - \frac{x^3}{3} \right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 5: Find the minimum area of triangle OPQ

To find the minimum area of triangle \( OPQ \), we need to minimize:
\[ A = \frac{(1 + \alpha^2)^2}{4\alpha} \]
We can find the critical points by taking the derivative and setting it to zero. After differentiating and solving, we find that the minimum occurs at \( \alpha = \frac{1}{\sqrt{3}} \).

### Step 6: Calculate the minimum area at \( \alpha = \frac{1}{\sqrt{3}} \)

Substituting \( \alpha = \frac{1}{\sqrt{3}} \) into the area formula:
\[ A_{min} = \frac{(1 + \frac{1}{3})^2}{4 \cdot \frac{1}{\sqrt{3}}} = \frac{\left(\frac{4}{3}\right)^2}{\frac{4}{\sqrt{3}}} = \frac{\frac{16}{9}}{\frac{4}{\sqrt{3}}} = \frac{4\sqrt{3}}{9} \]

### Step 7: Relate the areas to find \( k \)

We know:
\[ A_{min} = k \cdot A_{curve} \]
Substituting the areas:
\[ \frac{4\sqrt{3}}{9} = k \cdot \frac{2}{3} \]
Solving for \( k \):
\[ k = \frac{4\sqrt{3}}{9} \cdot \frac{3}{2} = \frac{2\sqrt{3}}{3} \]

### Final Answer

Thus, the value of \( k \) is:
\[ \boxed{\frac{2}{\sqrt{3}}} \]