Aspire Faculty ID #17944 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

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:

Solution

$y = \frac{1 - \alpha x}{2}$

Put this in equation of hyperbola

$x^2 - 9\left(\frac{1 - \alpha x}{2}\right)^2 = 9$

$\Rightarrow (4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$

Line does not intersect hyperbola

$\Rightarrow D < 0$

$\Rightarrow \alpha^2 < \frac{5}{9}$

$\Rightarrow \alpha \in \left(-\frac{\sqrt{5}}{3}, \frac{\sqrt{5}}{3}\right)$

$\Rightarrow \frac{\sqrt{5}}{3} \approx 0.74$

$\Rightarrow \alpha = 0.8$ (possible value)

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