JEE Main 2026 — 22 January, Morning Shift — Question 10

Given line: $\alpha x + 2y = 1 \Rightarrow y = \frac{1 - \alpha x}{2}$. Substitute into hyperbola: $x^2 - 9\left(\frac{1 - \alpha x}{2}\right)^2 = 9$. Simplify: $x^2 - \frac{9(1 - 2\alpha x + \alpha^2 x^2)}{4} = 9$. Multiply by 4: $4x^2 - 9(1 - 2\alpha x + \alpha^2 x^2) = 36$. $4x^2 - 9 + 18\alpha x - 9\alpha^2 x^2 = 36$. $(4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$. For no intersection, discriminant < 0: $(18\alpha)^2 - 4(4 - 9\alpha^2)(-45) < 0$. $324\alpha^2 + 180(4 - 9\alpha^2) < 0$. $324\alpha^2 + 720 - 1620\alpha^2 < 0 \Rightarrow -1296\alpha^2 + 720 < 0$. $1296\alpha^2 > 720 \Rightarrow \alpha^2 > \frac{720}{1296} = \frac{5}{9}$. Thus $\alpha \in (-\infty, -\frac{\sqrt{5}}{3}) \cup (\frac{\sqrt{5}}{3}, \infty)$. $\frac{\sqrt{5}}{3} \approx 0.745$. Among options, only 0.8 > 0.745. Hence possible value is $\alpha = 0.8$, which corresponds to option B.