The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If this line pasess through the point of intersection of the nearest directrix and the X-axis, then the eccentricity of the hyperbola is ___

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Solution

On substituting $(\frac{a}{e}, 0)$ in $y = - 2 x + 1$ , we get
$0 = - \frac{2 a}{e} + 1$
$\Rightarrow \frac{a}{e} = \frac{1}{2}$
Also, $y = - 2 x + 1$ is tangent to hyperbola.
$∴ 1 = 4 a^{2} - b^{2} \Rightarrow \frac{1}{a^{2}} = 4 - (e^{2} - 1) \Rightarrow \frac{4}{e^{2}} = 5 - e^{2}$
$\Rightarrow e^{4} - 5 e^{2} + 4 = 0 \Rightarrow (e^{2} - 4) (e^{2} - 1) = 0$
$\Rightarrow e = 2, e = 1$

e = 1 gives the conic as parabola. But conic is given as hyperbola, hence e = 2.

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