Given hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

For this standard hyperbola:

• Eccentricity: $e=\sqrt{1+\frac{b^2}{a^2}}$
• Directrices: $x=\pm \frac{a}{e}$

The nearest directrix to the origin on the right side is: $x=\frac{a}{e}$ Its intersection with the $x$-axis is the point $\left(\frac{a}{e},0\right)$

Use the condition that the line passes through this point

The line is $2x+y=1$

Since it passes through $\left(\frac{a}{e},0\right)$, $2\left(\frac{a}{e}\right)+0=1$ $\frac{2a}{e}=1$ $e=2a$

Use the tangent condition

Rewrite the line as $y=1-2x$

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ a line $y=mx+c$ is tangent if $c^2=a^2m^2-b^2$

Here, $m=-2,\quad c=1$ so $1=a^2(4)-b^2$ $4a^2-b^2=1$

Use the eccentricity relation

Since $e^2=1+\frac{b^2}{a^2}$ and from step 2, $e=2a$, we get $(2a)^2=1+\frac{b^2}{a^2}$ $4a^2=1+\frac{b^2}{a^2}$

Multiply by $a^2$: $4a^4=a^2+b^2$

From step 3, $b^2=4a^2-1$

Substitute: $4a^4=a^2+(4a^2-1)$ $4a^4=5a^2-1$ $4a^4-5a^2+1=0$

Let $t=a^2$. Then $4t^2-5t+1=0$ $(4t-1)(t-1)=0$ So, $t=1 \quad \text{or} \quad t=\frac14$

Check valid value using } e=2a

Since eccentricity of a hyperbola must satisfy $e>1$:

• If $a^2=1$, then $a=1$ and $e=2a=2$
• If $a^2=\frac14$, then $a=\frac12$ and $e=2a=1$, which is impossible for a hyperbola

Hence the valid value is $e=2$

Final answer

The eccentricity of the hyperbola is $\boxed{2}$