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

The line 2x + y = 1 is tangent to the hyperbola x²/a² - y²/b² = 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 $\text{Hyperbola: } \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1, \qquad \text{Line } 2x + y = 1 \text{ is tangent}$ $\text{The line passes through (nearest directrix)} \cap (x\text{-axis})$ $\text{To find: eccentricity } e \text{ of the hyperbola}$ $1.\ \text{tangency condition} \Rightarrow \text{relation between } a, b$ $2.\ \text{directrix's } x\text{-intercept} \Rightarrow a = e/2$ $3.\ e^2 = 1 + \dfrac{b^2}{a^2}; \text{ solve for } e$ $e = \sqrt{1 + \dfrac{b^2}{a^2}}, \qquad \text{directrices: } x = \pm \dfrac{a}{e}$ $2x + y = 1 \Rightarrow y = -2x + 1: \quad m = -2, \ c = 1$ $c^2 = a^2 m^2 - b^2 \Rightarrow 1 = 4a^2 - b^2$ $b^2 = 4a^2 - 1 \qquad \cdots (1)$ $\text{Directrix } x = \dfrac{a}{e} \ \text{ meets } x\text{-axis at } \left(\dfrac{a}{e}, 0\right)$ $\text{On line } (y=0): \ 2x = 1 \Rightarrow \dfrac{a}{e} = \dfrac{1}{2}$ $a = \dfrac{e}{2} \qquad \cdots (2)$ $e^2 = 1 + \dfrac{b^2}{a^2} = 1 + \dfrac{4a^2 - 1}{a^2} = 5 - \dfrac{1}{a^2}$ $a = \dfrac{e}{2} \Rightarrow \dfrac{1}{a^2} = \dfrac{4}{e^2}$ $e^2 = 5 - \dfrac{4}{e^2} \Rightarrow e^4 - 5e^2 + 4 = 0$ $(e^2 - 1)(e^2 - 4) = 0 \Rightarrow e^2 = 1 \text{ or } 4$ $e > 1 \Rightarrow e^2 = 4 \Rightarrow e = 2$ $e = 2 \Rightarrow a = 1, \ b^2 = 3: \ x^2 - \dfrac{y^2}{3} = 1$ $\text{Directrix } x = \dfrac{1}{2}: \ 2\left(\dfrac{1}{2}\right) + 0 = 1 \ \text{✓}$ $\boxed{e = 2}$