The line 2x + y = 1 is tangent to the hyperbo | vedclass

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(B) The equation of the line is $y = -2x + 1$,so the slope $m = -2$.The nearest directrix is $x = \frac{a}{e}$. The point of intersection of the direc

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$2$

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(B) The equation of the line is $y = -2x + 1$,so the slope $m = -2$.The nearest directrix is $x = \frac{a}{e}$. The point of intersection of the directrix and the $x$-axis is $(\frac{a}{e}, 0)$.Since the line passes through $(\frac{a}{e}, 0)$,we have $0 = -2(\frac{a}{e}) + 1$,which implies $\frac{2a}{e} = 1$,or $a = \frac{e}{2}$.The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Here $c = 1$ and $m = -2$,so $1^2 = a^2(-2)^2 - b^2$,which gives $1 = 4a^2 - b^2$.Substitute $a^2 = \frac{e^2}{4}$ into the equation: $1 = 4(\frac{e^2}{4}) - b^2$,so $1 = e^2 - b^2$,which means $b^2 = e^2 - 1$.We also know for a hyperbola that $b^2 = a^2(e^2 - 1)$.Substituting $b^2 = e^2 - 1$ into this,we get $e^2 - 1 = a^2(e^2 - 1)$.Since $e > 1$,$e^2 - 1 
eq 0$,so $a^2 = 1$,which means $a = 1$.Since $a = \frac{e}{2}$,we have $1 = \frac{e}{2}$,so $e = 2$.

The line $2x + y = 1$ 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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