Question

The line $ 2x+y=1 $ is tangent to the hyperbola $ \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ . If this line passes through the point of intersection of the nearest directrix and X-axis, then the eccentricity of the hyperbola is
A. 2
B. 1
C. $ \sqrt{2} $
D. 4

Answer 1

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.