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**Hint:**We first use the condition of tangency of a line $ y=mx+c $ for hyperbola $ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ . Then we use the coordinates of point of intersection of the nearest directrix and X-axis to find the relation between a and e. we have the relation of eccentricity as $ {{b}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right) $ . We use these three equations to solve and get the values of the unknowns.

**Complete step by step answer:**

The line $ 2x+y=1 $ is tangent to the hyperbola $ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ .

The equation of directrix the hyperbola $ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ is $ x=\pm \dfrac{a}{e} $ .

This line $ 2x+y=1 $ passes through the point of intersection of the nearest directrix and X-axis.

Now we do not have the exact value of the terms a and e. so, without loss of generality we are taking the value of $ x=\dfrac{a}{e} $ for the nearest directrix.

The point of intersection of the nearest directrix and X-axis is \[\left( \dfrac{a}{e},0 \right)\]. This point satisfies the line $ 2x+y=1 $.

Putting the values, we get \[\dfrac{2a}{e}=1\Rightarrow e=2a\].

Now the slope of the line $ 2x+y=1 $ will be equal to the slope of the tangent of $ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ .

We know that for hyperbola $ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , condition of tangency of a line $ y=mx+c $ with m being the slope is $ {{c}^{2}}={{a}^{2}}m-{{b}^{2}} $ .

Now for line $ 2x+y=1\Rightarrow y=-2x+1 $ , we have $ m=-2,c=1 $ which gives $ 1=4{{a}^{2}}-{{b}^{2}} $ .

We also have the relation of eccentricity as $ {{b}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right) $ .

We have in total three equations and three unknowns.

They are \[e=2a;1=4{{a}^{2}}-{{b}^{2}};{{b}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right)\].

From the last two equations we replace b and get

\[\begin{align}

& \Rightarrow 4{{a}^{2}}-{{a}^{2}}\left( {{e}^{2}}-1 \right)=1 \\

& \Rightarrow {{a}^{2}}\left( 5-{{e}^{2}} \right)=1 \\

\end{align}\]

Now we have $ a=\dfrac{e}{2} $ which we place in the equation \[{{a}^{2}}\left( 5-{{e}^{2}} \right)=1\].

\[\begin{align}

& \dfrac{{{e}^{2}}}{4}\left( 5-{{e}^{2}} \right)=1 \\

& \Rightarrow {{e}^{4}}-5{{e}^{2}}+4=0 \\

\end{align}\]

We apply factorisation and get \[\left( {{e}^{2}}-4 \right)\left( {{e}^{2}}-1 \right)=0\].

From this we get 4 values of e which are $ e=\pm 2,\pm 1 $ . But we know that in the case of hyperbola the value of e is always $ e>1 $ . This eliminates three possible values of e from $ e=\pm 2,\pm 1 $ .

Therefore, the value of e is $ e=2 $ .

**The correct option is A**.

**Note:**

We need to remember the conditions of the eccentricity of the hyperbolas and the other conics. The conditions and the equations would have been the same for the point $ x=-\dfrac{a}{e} $ . The given line is closer to the directrix depending on the value of e.

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