Question

If the line \( ax + 2y = 1 \), where \( a \in \mathbb{R} \), does not meet the hyperbola \( x^2 - 9y^2 = 9 \), then a possible value of \( a \) is:

• A. 0.5
• B. 0.6
• C. 0.8
• D. 0.7

Hint:

When a line does not intersect a conic, substitute the line into the conic equation and ensure the resulting quadratic has a negative discriminant.

Answer 1

The Correct Option is D

To determine the value of \( a \) such that the line \( ax + 2y = 1 \) does not intersect the hyperbola \( x^2 - 9y^2 = 9 \), we need to check for conditions where the line is entirely outside the hyperbola. This occurs when the quadratic equation obtained by eliminating \( y \) has no real solutions.

First, let's substitute \( y = \frac{1 - ax}{2} \) from the line equation into the hyperbola equation:

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

Simplifying this,

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

\(x^2 - \frac{9}{4}(1 - 2ax + a^2x^2) = 9\)

\(x^2 - \frac{9}{4} + \frac{9ax}{2} - \frac{9a^2x^2}{4} = 9\)

Rearranging terms gives us:

\(4x^2 - 9a^2x^2 + 18ax - 9 - 36 = 0\)

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

This is a quadratic equation in \( x \). For this quadratic to have no real solutions, the discriminant must be less than zero.

The discriminant \(\Delta\) for a quadratic equation \(Ax^2 + Bx + C = 0\) is given by \(\Delta = B^2 - 4AC\).

For our equation, \(A = 4 - 9a^2\), \(B = 18a\), \(C = -45\). Thus:

\(\Delta = (18a)^2 - 4(4 - 9a^2)(-45)\)

\(= 324a^2 + 180(4 - 9a^2)\)

\(= 324a^2 + 720 - 1620a^2\)

\(= 720 - 1296a^2\)

We need \( \Delta < 0 \):

\(720 - 1296a^2 < 0\)

\(720 < 1296a^2\)

\(\frac{720}{1296} < a^2\)

\(\frac{5}{9} < a^2\)

Thus, \( a^2 > \frac{5}{9} \), implying \( a > \sqrt{\frac{5}{9}} \approx 0.745 \) or \( a < -\sqrt{\frac{5}{9}} \approx -0.745 \).

Given options are positive, thus \( a \approx 0.7 \) is the closest valid answer.

Therefore, a possible value of \( a \) for which the line does not meet the hyperbola is 0.7.