If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^{2} -

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If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^{2} - 9y^{2} = 9$, then a possible value of $\alpha$ is :

Accepted Answer

$y = \frac{1 - \alpha x}{2}$Put this in equation of hyperbola$${	herefore x^{2} - 9\left( \frac{1 - \alpha x}{2} ight)^{2} = 9 }{\left( 4 - 9\alpha^{2} ight)x^{2} + 18\alpha x - 45 = 0 }$$∵ line does not intersect hyperbola$${	herefore D < 0 }{\Rightarrow \alpha^{2} - \frac{5}{9} > 0 }{\Rightarrow \alpha \in \left( - \infty, - \frac{\sqrt{5}}{3} ight) \cup \left( \frac{\sqrt{5}}{3},\infty ight) }$$Here $\frac{\sqrt{5}}{3} \simeq 0.74$

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