Q31 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
If the line $\alpha x+2 y=1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^{2}-9 y^{2}=9$, then a possible value of $\alpha$ is:
- A0.5
- B0.6
- C0.7
- Correct answerD0.8
Solution
Hyperbola: $\frac{x^2}{9} - y^2 = 1$. Line: $y = \frac{1 - \alpha x}{2}$.
Substituting: $x^2 - 9 \cdot \frac{(1-\alpha x)^2}{4} = 9$
$\Rightarrow (4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$.
For no intersection (when $4 - 9\alpha^2 \neq 0$): discriminant $< 0$.
$D = 324\alpha^2 + 180(4 - 9\alpha^2) < 0$
$\Rightarrow -1296\alpha^2 + 720 < 0 \Rightarrow \alpha^2 > \frac{5}{9}$.
$|\alpha| > \frac{\sqrt{5}}{3} \approx 0.745$.
Among the options, only $\alpha = 0.8$ satisfies this condition.
Substituting: $x^2 - 9 \cdot \frac{(1-\alpha x)^2}{4} = 9$
$\Rightarrow (4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0$.
For no intersection (when $4 - 9\alpha^2 \neq 0$): discriminant $< 0$.
$D = 324\alpha^2 + 180(4 - 9\alpha^2) < 0$
$\Rightarrow -1296\alpha^2 + 720 < 0 \Rightarrow \alpha^2 > \frac{5}{9}$.
$|\alpha| > \frac{\sqrt{5}}{3} \approx 0.745$.
Among the options, only $\alpha = 0.8$ satisfies this condition.
Hence, the correct answer is option: 'D'.