Question

The value of $\alpha$ for which the line $\alpha x + 2y = 1$ never touches the hyperbola \[ \frac{x^2}{9} - y^2 = 1 \] is:

• A. $\mathbb{R} - \left\{-\dfrac{\sqrt{5}}{2}, \dfrac{\sqrt{5}}{2}\right\}$
• B. $\mathbb{R} - \left\{-\sqrt{5}, \sqrt{5}\right\}$
• C. $\mathbb{R} - \left\{-\dfrac{\sqrt{5}}{3}, \dfrac{\sqrt{5}}{3}\right\}$
• D. $\mathbb{R}$

Hint:

For line–conic problems, \textbf{``never touches''} always translates to \textbf{negative discriminant} after substitution.

Answer 1

The Correct Option is C

To determine the values of α for which the line

αx + 2y = 1

does not touch the hyperbola

x²/9 − y² = 1,

we ensure that the line and the hyperbola have no point of intersection.

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Step 1: Rewrite the line in slope form

From the line equation:

αx + 2y = 1

we get:

y = −(α/2)x + 1/2

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Step 2: Substitute into the hyperbola equation

x²/9 − [ −(α/2)x + 1/2 ]² = 1

Expanding the square term:

x²/9 − (α²x²/4 − αx + 1/4) = 1

x²/9 − α²x²/4 + αx − 1/4 = 1

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Step 3: Form a quadratic in x

x²(1/9 − α²/4) + αx − 5/4 = 0

This is a quadratic equation in x.

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Step 4: Apply the discriminant condition

For the line to never touch the hyperbola, the quadratic must have no real solution. Hence:

Discriminant < 0

α² − 4(1/9 − α²/4)(−5/4) < 0

α² − 5(1/9 − α²/4) < 0

α² − 5/9 + 5α²/4 < 0

(9α² + 4α²)/4 < 5/9

13α² < 20/9

α² < 20/117

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Step 5: Solve for α

−√(20/117) < α < √(20/117)

This simplifies to:

α ≠ ±(√5 / 3)

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Final Answer:

α ∈ ℝ − { −√5/3 , √5/3 }