Question

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

• A. \( R - \sqrt{\frac{5}{2}} \)
• B. \( R - \sqrt{5}, \sqrt{5} \)
• C. \( R - \sqrt{\frac{5}{3}} \)
• D. \( R \)

Hint:

When solving geometry problems involving conic sections and lines, use the distance formula from a point to a line to determine the condition for intersection or non-intersection.

Answer 1

The Correct Option is C

To find the value(s) of \( \alpha \) for which the line \( \alpha x + 2y = 1 \) never touches the hyperbola \(\frac{x^2}{9} - \frac{y^2}{1} = 1\), we need to determine when there are no points of intersection between the line and the hyperbola.

The equation of the hyperbola is given in the standard form: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a^2 = 9\) and \(b^2 = 1\).

First, let's substitute \(y\) from the line equation into the hyperbola equation. Solving for \(y\) in terms of \(x\) gives:

\(y = \frac{1 - \alpha x}{2}\)

Substituting this into the hyperbola equation, we get:

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

Simplify and solve the equation:

\(\frac{x^2}{9} - \frac{(1 - \alpha x)^2}{4} = 1\)

Multiply the entire equation by 36 to eliminate the fractions:

\(4x^2 - 9(1 - 2\alpha x + \alpha^2 x^2) = 36\)

Expanding the equation gives:

\(4x^2 - 9 + 18\alpha x - 9\alpha^2 x^2 = 36\)

Combining like terms:

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

This is a quadratic equation in terms of \(x\). For the line to never touch the hyperbola, the discriminant of this quadratic must be less than zero:

\(D = (18\alpha)^2 - 4(4 - 9\alpha^2)(-45) < 0\)

Simplify the expression for the discriminant:

\(324\alpha^2 + 720 - 1620\alpha^2 < 0\)

Simplifying further gives:

\(1296\alpha^2 - 720 > 0\)

Divide the entire inequality by 36:

\(36\alpha^2 - 20 > 0\)

Simplifying further:

\(\alpha^2 > \frac{5}{9}\)

This implies:

\(|\alpha| > \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\)

The range of \(\alpha\) is:

\(\alpha > \frac{\sqrt{5}}{3} \quad \text{or} \quad \alpha < -\frac{\sqrt{5}}{3}\)

Therefore, the line does not touch the hyperbola when \(|\alpha| \gt \frac{\sqrt{5}}{3}\). Thus the value(s) of \(\alpha\) are in \(R - \frac{\sqrt{5}}{3}\).

The correct answer is:

\( R - \sqrt{\frac{5}{3}} \)