The line 2x+y=1 is tangent to the hyperbola (\frac{x^2}{a^2} - \frac{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?

Equation of tangent is 2x+y=1
The equation of the hyperbola is (\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1)
The equation of the tangent to the hyperbola is given by (yy_1 - \frac{x x_1}{a^2} + \frac{y_1^2}{b^2} -1 = 0)
Comparing with 2x+y-1=0, we have
(\frac{x_1}{a^2} = -2), (\frac{y_1}{b^2} = 1), and (\frac{y_1^2}{b^2} -1 = -1)
Therefore, (x_1 = -2a^2) and (y_1 = b^2)
Since ((x_1, y_1)) lies on the hyperbola, we have (\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1)
(\frac{(-2a^2)^2}{a^2} - \frac{(b^2)^2}{b^2} = 1)
(4a^2 - b^2 = 1)
The nearest directrix is given by x = (\frac{a}{e})
The point of intersection of the directrix and the x-axis is ((\frac{a}{e}, 0))
The line 2x+y=1 passes through ((\frac{a}{e}, 0))
Therefore, (2(\frac{a}{e}) + 0 = 1)
(\frac{2a}{e} = 1)
(e = 2a)
Since (b^2 = a^2(e^2 - 1)), we have
(4a^2 - a^2(e^2 - 1) = 1)
(4a^2 - a^2(4a^2 - 1) = 1)
(4a^2 - 4a^4 + a^2 = 1)
(5a^2 - 4a^4 = 1)
(4a^4 - 5a^2 + 1 = 0)
Let (x = a^2)
(4x^2 - 5x + 1 = 0)
(4x^2 - 4x - x + 1 = 0)
(4x(x - 1) - 1(x - 1) = 0)
((4x - 1)(x - 1) = 0)
(x = 1) or (x = \frac{1}{4})
(a^2 = 1) or (a^2 = \frac{1}{4})
If (a^2 = 1), then (a = 1)
(e = 2a = 2)
If (a^2 = \frac{1}{4}), then (a = \frac{1}{2})
(e = 2a = 1)
Since the eccentricity is greater than 1, e=2