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

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(B) The line $2x + y = 1$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The equation of the directrix is $x = \frac{a}{e}$. T

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$2$

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(B) The line $2x + y = 1$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The equation of the directrix is $x = \frac{a}{e}$. The line passes through the point $(\frac{a}{e}, 0)$,so $2(\frac{a}{e}) + 0 = 1$,which gives $2a = e$. The condition for the line $y = mx + c$ to be a tangent to the hyperbola is $c^2 = a^2m^2 - b^2$. Here,$y = -2x + 1$,so $m = -2$ and $c = 1$. Thus,$1^2 = a^2(-2)^2 - b^2$,which simplifies to $4a^2 - b^2 = 1$. We know that for a hyperbola,$b^2 = a^2(e^2 - 1)$. Substituting $b^2$ into the tangent condition: $4a^2 - a^2(e^2 - 1) = 1$. Since $e = 2a$,we have $a = \frac{e}{2}$,so $a^2 = \frac{e^2}{4}$. Substituting $a^2$: $4(\frac{e^2}{4}) - \frac{e^2}{4}(e^2 - 1) = 1$. $e^2 - \frac{e^4}{4} + \frac{e^2}{4} = 1$. Multiply by $4$: $4e^2 - e^4 + e^2 = 4$. $e^4 - 5e^2 + 4 = 0$. $(e^2 - 4)(e^2 - 1) = 0$. Since $e > 1$ for a hyperbola,$e^2 = 4$,so $e = 2$.

The line $2x + y = 1$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $(a > b)$. If this line passes through the point of intersection of a directrix and the positive $X$-axis,then the eccentricity of that hyperbola is

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