Question

If one of the lines given by the pair of lines \( 3x^2 - 2y^2 + axy = 0 \) is making an angle \( 60^\circ \) with the x-axis, then \( a = \)

• A. \( \sqrt{3} \)
• B. \( \frac{1}{\sqrt{3}} \)
• C. \( 3 \)
• D. \( \frac{1}{3} \)

Hint:

To find the value of the coefficient \( a \) when a line from a homogeneous second-degree equation makes a known angle with the x-axis, substitute the slope into the equation and solve for \( a \).

Answer 1

The Correct Option is A

The given equation is a homogeneous equation representing a pair of straight lines: \[ 3x^2 - 2y^2 + axy = 0 \] For such an equation, the angle \( \theta \) between the lines is given by: \[ \tan\theta = \left| \frac{a}{b - c} \right| \] where the general form is \( ax^2 + 2hxy + by^2 = 0 \). Comparing: - \( a = 3 \), \( b = -2 \), \( 2h = a \Rightarrow h = \frac{a}{2} \) Now apply the formula: \[ \tan\theta = \left| \frac{a}{3 + 2} \right| = \left| \frac{a}{5} \right| \] But the angle is given as \( 60^\circ \), so: \[ \tan 60^\circ = \sqrt{3} = \left| \frac{a}{5} \right| \Rightarrow a = 5\sqrt{3} \] Wait, this contradicts the correct answer marked as \( \sqrt{3} \). Let's re-check using the right form. Actually, the correct angle formula for the pair of lines \( ax^2 + 2hxy + by^2 = 0 \) is: \[ \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right| \] From the given equation: - \( a = 3 \), \( b = -2 \), \( h = \frac{a}{2} \) Now apply: \[ \tan 60^\circ = \sqrt{3} = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right| = \left| \frac{2\sqrt{ \left( \frac{a}{2} \right)^2 - 3(-2) }}{3 - 2} \right| = 2\sqrt{ \frac{a^2}{4} + 6 } \] Set this equal to \( \sqrt{3} \): \[ \sqrt{3} = 2\sqrt{ \frac{a^2}{4} + 6 } \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{ \frac{a^2}{4} + 6 } \Rightarrow \frac{3}{4} = \frac{a^2}{4} + 6 \Rightarrow \frac{3}{4} - 6 = \frac{a^2}{4} \Rightarrow \frac{-21}{4} = \frac{a^2}{4} \] This gives imaginary result — a contradiction, suggesting earlier confusion. Let’s simplify. Instead, rewrite the given equation in terms of slopes: Assume line is of form: \[ y = mx \Rightarrow \text{substitute } y = mx \text{ into } 3x^2 - 2y^2 + axy = 0 \Rightarrow 3x^2 - 2m^2x^2 + a x . mx = 0 \Rightarrow (3 - 2m^2 + am)x^2 = 0 \Rightarrow 3 - 2m^2 + am = 0 \] Given the line makes \( 60^\circ \) with x-axis, \( m = \tan(60^\circ) = \sqrt{3} \) Substitute: \[ 3 - 2(\sqrt{3})^2 + a(\sqrt{3}) = 0 \Rightarrow 3 - 6 + a\sqrt{3} = 0 \Rightarrow -3 + a\sqrt{3} = 0 \Rightarrow a = \sqrt{3} \]