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# 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. 2B. 1C. $\sqrt{2}$ D. 4

Last updated date: 12th Aug 2024
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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.

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.