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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
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Verified Answer
The correct answer is:
2
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 $e=2$
$$
\begin{aligned}
0 & =-\frac{2 a}{e}+1 \\
\Rightarrow \quad \frac{a}{e} & =\frac{1}{2}
\end{aligned}
$$
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 $e=2$
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