The chrods of contact of the pair of tangents to the circle `x^(2)+y^(2)=1` dravwm from any point on the line 2x+y=4 paas through the point `(alpha,beta)` then find `alpha and beta`.

Answer

To solve the problem, we need to find the values of \( \alpha \) and \( \beta \) such that the chords of contact of the pair of tangents to the circle \( x^2 + y^2 = 1 \) drawn from any point on the line \( 2x + y = 4 \) pass through the point \( (\alpha, \beta) \).

### Step-by-Step Solution:

1. **Identify the Circle and Line**:
The given circle is \( x^2 + y^2 = 1 \) and the line is \( 2x + y = 4 \).

2. **Equation of the Chord of Contact**:
The equation of the chord of contact from a point \( (x_1, y_1) \) to the circle \( x^2 + y^2 = r^2 \) is given by:
\[ x x_1 + y y_1 = r^2 \]
For our circle, \( r^2 = 1 \), so the equation becomes:
\[ x x_1 + y y_1 = 1 \]

3. **Parameterize the Point on the Line**:
Let \( (x_1, y_1) \) be a point on the line \( 2x + y = 4 \). We can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = 4 - 2x_1 \]

4. **Substituting into the Chord of Contact Equation**:
Substitute \( y_1 \) into the chord of contact equation:
\[ x x_1 + y (4 - 2x_1) = 1 \]
Rearranging gives:
\[ x x_1 + 4y - 2x_1y = 1 \]
This can be rewritten as:
\[ x x_1 - 2x_1y + 4y = 1 \]

5. **Comparing with the Given Line**:
We need to find the values of \( \alpha \) and \( \beta \) such that the chord of contact passes through \( (\alpha, \beta) \). The equation can be rearranged to:
\[ \alpha x_1 - 2x_1 \beta + 4\beta = 1 \]
This implies:
\[ x_1(\alpha - 2\beta) + 4\beta = 1 \]

6. **Finding Relationships**:
For this equation to hold for all \( x_1 \), the coefficients must equal zero. Thus, we have:
\[ \alpha - 2\beta = 0 \quad \text{(1)} \]
and
\[ 4\beta = 1 \quad \text{(2)} \]

7. **Solving the Equations**:
From equation (2):
\[ \beta = \frac{1}{4} \]
Substitute \( \beta \) into equation (1):
\[ \alpha - 2\left(\frac{1}{4}\right) = 0 \implies \alpha - \frac{1}{2} = 0 \implies \alpha = \frac{1}{2} \]

8. **Final Values**:
Thus, the values of \( \alpha \) and \( \beta \) are:
\[ \alpha = \frac{1}{2}, \quad \beta = \frac{1}{4} \]

### Conclusion:
The final answer is:
\[ (\alpha, \beta) = \left(\frac{1}{2}, \frac{1}{4}\right) \]