The line `2x + y = 1` is tangent to the hyperbola `x^2/a^2-y^2/b^2=1`. If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

Explanation

To solve the problem step by step, we will follow the given information about the hyperbola and the tangent line.

### Step 1: Identify the equation of the hyperbola and the tangent line
The equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
The equation of the tangent line is:
\[ 2x + y = 1 \]

### Step 2: Find the condition for the tangent
For a line \( y = mx + c \) to be a tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the condition is given by:
\[ c^2 = a^2m^2 - b^2 \]
Here, we can rewrite the tangent line in the form \( y = -2x + 1 \), where \( m = -2 \) and \( c = 1 \).

Substituting these values into the condition:
\[ 1^2 = a^2(-2)^2 - b^2 \]
This simplifies to:
\[ 1 = 4a^2 - b^2 \quad \text{(Equation 1)} \]

### Step 3: Find the point of intersection of the nearest directrix and the x-axis
The directrix of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by \( x = \frac{a}{e} \), where \( e \) is the eccentricity of the hyperbola.

The nearest directrix intersects the x-axis at the point:
\[ \left(-\frac{a}{e}, 0\right) \]

### Step 4: Substitute the point into the tangent line equation
Since the line \( 2x + y = 1 \) passes through the point of intersection, we substitute \( x = -\frac{a}{e} \) and \( y = 0 \) into the tangent line equation:
\[ 2\left(-\frac{a}{e}\right) + 0 = 1 \]
This simplifies to:
\[ -\frac{2a}{e} = 1 \]
Multiplying both sides by -1 gives:
\[ \frac{2a}{e} = -1 \]
Thus:
\[ 2a = e \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
Now we have two equations:
1. \( 1 = 4a^2 - b^2 \)
2. \( 2a = e \)

From Equation 2, we can express \( a \) in terms of \( e \):
\[ a = \frac{e}{2} \]

### Step 6: Substitute \( a \) into Equation 1
Substituting \( a = \frac{e}{2} \) into Equation 1:
\[ 1 = 4\left(\frac{e}{2}\right)^2 - b^2 \]
This simplifies to:
\[ 1 = 4 \cdot \frac{e^2}{4} - b^2 \]
\[ 1 = e^2 - b^2 \]
Rearranging gives:
\[ b^2 = e^2 - 1 \quad \text{(Equation 3)} \]

### Step 7: Find the eccentricity \( e \)
The eccentricity \( e \) of the hyperbola is defined as:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \( b^2 \) from Equation 3 and \( a^2 = \left(\frac{e}{2}\right)^2 = \frac{e^2}{4} \):
\[ e = \sqrt{1 + \frac{e^2 - 1}{\frac{e^2}{4}}} \]
This simplifies to:
\[ e = \sqrt{1 + \frac{4(e^2 - 1)}{e^2}} = \sqrt{1 + 4 - \frac{4}{e^2}} = \sqrt{5 - \frac{4}{e^2}} \]

### Step 8: Solve for \( e \)
Squaring both sides gives:
\[ e^2 = 5 - \frac{4}{e^2} \]
Multiplying through by \( e^2 \) leads to:
\[ e^4 = 5e^2 - 4 \]
Rearranging gives:
\[ e^4 - 5e^2 + 4 = 0 \]
Let \( x = e^2 \), then we have:
\[ x^2 - 5x + 4 = 0 \]
Factoring gives:
\[ (x - 4)(x - 1) = 0 \]
Thus, \( x = 4 \) or \( x = 1 \). Therefore, \( e^2 = 4 \) or \( e^2 = 1 \).

Since \( e \) must be greater than 1 for a hyperbola, we take:
\[ e = 2 \]

### Final Answer
The eccentricity of the hyperbola is \( \boxed{2} \).