Eccentricity of Hyperbola Tangent to Given Line

To find the eccentricity of the hyperbola, we follow these steps:

1. Condition of Tangency:
   The equation of the hyperbola is:
   $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
   The equation of the given tangent line is:
   $2x + y = 1 \implies y = -2x + 1$
   Here, the slope of the line is $m = -2$ and the $y$-intercept is $c = 1$.

For the line $y = mx + c$ to be tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, it must satisfy the condition:
$c^2 = a^2m^2 - b^2$
Substituting $m = -2$ and $c = 1$:
$1^2 = a^2(-2)^2 - b^2$
$1 = 4a^2 - b^2 \quad \text{--- (1)}$

1. Intersection with the Nearest Directrix:
   The point of contact $(x_0, y_0)$ of the tangent line with the hyperbola is given by:
   $x_0 = -\frac{a^2m}{c} = -\frac{a^2(-2)}{1} = 2a^2 > 0$
   Since the $x$-coordinate of the point of contact is positive, the point of contact lies on the right branch ($x > 0$) of the hyperbola.

Therefore, the nearest directrix to this branch is:
$x = \frac{a}{e}$
The intersection of this directrix with the $x$-axis is the point $\left(\frac{a}{e}, 0\right)$.

Since the tangent line $2x + y = 1$ passes through $\left(\frac{a}{e}, 0\right)$:
$2\left(\frac{a}{e}\right) + 0 = 1 \implies a = \frac{e}{2} \quad \text{--- (2)}$

1. Determining the Eccentricity $e$:
   For a hyperbola, the relation between $a$, $b$, and the eccentricity $e$ is:
   $b^2 = a^2(e^2 - 1)$

Substituting this into equation (1):
$1 = 4a^2 - a^2(e^2 - 1)$
$1 = a^2(5 - e^2)$

Now, substitute $a = \frac{e}{2}$ from equation (2):
$1 = \left(\frac{e}{2}\right)^2 (5 - e^2)$
$4 = e^2(5 - e^2)$
$e^4 - 5e^2 + 4 = 0$

Factoring this quadratic equation in terms of $e^2$:
$(e^2 - 1)(e^2 - 4) = 0$

This yields:
$e^2 = 1 \quad \text{or} \quad e^2 = 4$

Since the eccentricity of a hyperbola must be strictly greater than $1$ ($e > 1$), we discard $e^2 = 1$:
$e^2 = 4 \implies e = 2$

Thus, the eccentricity of the hyperbola is 2.