If the line alpha x+2y=1, where alpha in math | vedclass

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(B) The equation of the line is $y = \frac{1-\alpha x}{2}$.Substituting this into the hyperbola equation $x^2 - 9y^2 = 9$:$x^2 - 9\left(\frac{1-\alpha

Vedclass · Accepted Answer

$0.8$

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(B) The equation of the line is $y = \frac{1-\alpha x}{2}$.Substituting this into the hyperbola equation $x^2 - 9y^2 = 9$:$x^2 - 9\left(\frac{1-\alpha x}{2}\right)^2 = 9$$x^2 - \frac{9}{4}(1 - 2\alpha x + \alpha^2 x^2) = 9$$4x^2 - 9 + 18\alpha x - 9\alpha^2 x^2 = 36$$(4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$Since the line does not intersect the hyperbola, the discriminant $D $D = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) $324\alpha^2 + 180(4 - 9\alpha^2) $324\alpha^2 + 720 - 1620\alpha^2 $-1296\alpha^2 + 720 $1296\alpha^2 > 720$$\alpha^2 > \frac{720}{1296} = \frac{5}{9}$$|\alpha| > \frac{\sqrt{5}}{3} \approx 0.745$Comparing with the options, the only value greater than $0.745$ is $0.8$.

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:

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