If the line α x+2 y=1 , where α ∈ ℝ , does not meet the hyperbola x^ 2 -9 y^ 2 =9 , then a possible value of α is:

Correct answer: D. 0.8 Hyperbola: (x^2)/(9) - y^2 = 1 . Line: y = (1 - α x)/(2) . Substituting: x^2 - 9 · ((1-α x)^2)/(4) = 9 (4 - 9α^2)x^2 + 18α x - 45 = 0 . For no intersection (when 4 - 9α^2 ≠ 0 ): discriminant D = 324α^2 + 180(4 - 9α^2) -1296α^2 + 720 (5)/(9) . |α| > 5 3 0.745 . Among the options, only α = 0.8 satisfies this condition.

JEE Main 202622 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Hyperbola

If the line α x+2 y=1 , where α ∈ ℝ , does not meet the hyperbola...

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:

Options

A0.5
B0.6
C0.7
D0.8

Correct answer

D. 0.8

Step-by-step 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.

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