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The line $2 x + y = 1$ is tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ This line passes through point of intersection of nearest directrix and x-axis. Then the eccentricity of Hyperbola is

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a

$\sqrt{2}$

b

2

c

$\frac{3}{2}$

d

$\frac{3}{\sqrt{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} 2 x + y = 1 cuts x -axis at (\frac{1}{2}, 0) \\ \Rightarrow \frac{a}{e} = \frac{1}{2} \Rightarrow e = 2 a \end{array}$
$2 x + y = 1$ is tangent to hyperbola
$\begin{array}{l} \Rightarrow 1 = 4 a^{2} - b^{2} \\ 1 = 4 a^{2} - a^{2} (e^{2} - 1) \\ = 4 a^{2} - a^{2} (4 a^{2} - 1) \\ = 4 a^{2} - 4 a^{4} + a^{2} \\ \Rightarrow 4 a^{4} - 5 a^{2} + 1 = 0 \\ a^{2} = 1, \frac{1}{4} \end{array}$
$a^{2} = \frac{1}{4} \Rightarrow e = 1,$ it is not a hyperbola
Then, $a^{2} = 1 ⇒ e = 2$

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The line 2x+y=1 is tangent to the hyperbola x2a2−y2b2=1 This line passes through point of intersection of nearest directrix and x-axis. Then the eccentricity of Hyperbola is