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 passes through the point of intersection of the nearest directrix and the $x$-axis, then the eccentricity of the hyperbola is

Solution

On substituting $\left(\frac{a}{e}, 0\right)$ in $y=-2 x+1$, we get $$ \begin{aligned} 0 & =-\frac{2 a}{e}+1 \\ \Rightarrow \quad \frac{a}{e} & =\frac{1}{2} \end{aligned} $$ Also, $y=-2 x+1$ is tangent to hyperbola $$ \begin{array}{ll} \therefore & 1=4 a^2-b^2 \\ \Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right) \end{array} $$
$$ \begin{array}{ll} \Rightarrow & \frac{4}{e^2}=5-e^2 \\ \Rightarrow & e^4-5 e^2+4=0 \\ \Rightarrow & \left(e^2-4\right)\left(e^2-1\right)=0 \\ \Rightarrow & e=2, e=1 \end{array} $$ $e=1$ gives the conic as parabola. But conic is given as hyperbola, hence $e=2$

Hence, the correct answer is 2.