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

Concept: A straight line never touches a conic section if, after substitution, the resulting quadratic equation has no real solution. This happens when the discriminant is negative: \[ \Delta<0 \] Key idea used:
Substitute the equation of the line into the equation of the hyperbola
Analyze the discriminant of the resulting quadratic equation
Step 1: Express $y$ from the line equation. \[ \alpha x + 2y = 1 \quad \Rightarrow \quad y = \frac{1 - \alpha x}{2} \]
Step 2: Substitute $y$ into the hyperbola equation. \[ \frac{x^2}{9} - \left(\frac{1 - \alpha x}{2}\right)^2 = 1 \]
Step 3: Simplify the equation. \[ \frac{x^2}{9} - \frac{(1 - \alpha x)^2}{4} = 1 \] Multiply throughout by $36$ to remove denominators: \[ 4x^2 - 9(1 - 2\alpha x + \alpha^2 x^2) = 36 \]
Step 4: Expand and arrange terms. \[ 4x^2 - 9 + 18\alpha x - 9\alpha^2 x^2 - 36 = 0 \] \[ (4 - 9\alpha^2)x^2 + 18\alpha x - 45 = 0 \]
Step 5: Apply the discriminant condition. \[ \Delta = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) \] \[ \Delta = 324\alpha^2 + 180(4 - 9\alpha^2) \] \[ \Delta = 720 - 1296\alpha^2 \]
Step 6: For the line to never touch the hyperbola: \[ \Delta<0 \] \[ 720 - 1296\alpha^2<0 \] \[ \alpha^2>\frac{5}{9} \]
Step 7: Write the final set of values. \[ |\alpha|>\frac{\sqrt{5}}{3} \] Hence, \[ \alpha \in \mathbb{R} - \left\{-\frac{\sqrt{5}}{3}, \frac{\sqrt{5}}{3}\right\} \]