Question

If a pair of line given by \( (x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha \) are perpendicular. What is the value of α?

Answer 1

Step 1: Start with the equation: \[ (x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha \] Expanding both sides: \[ x^2 \cos^2 \alpha + 2xy \sin \alpha \cos \alpha + y^2 \sin^2 \alpha = x^2 \sin^2 \alpha + y^2 \sin^2 \alpha \] Simplifying: \[ x^2 \cos(2\alpha) + xy \sin(2\alpha) = 0 \]

Step 2: Substitute \( y = mx \) into the equation: \[ x^2 \cos(2\alpha) + x(mx) \sin(2\alpha) = 0 \] Simplifying: \[ \cos(2\alpha) + m \sin(2\alpha) = 0 \] Solving for \( m \): \[ m = -\cot(2\alpha) \]

Step 3: Apply the condition for perpendicular lines: \[ m_1 m_2 = -1 \quad \Rightarrow \quad \tan(2\alpha) = -1 \] Solving for \( \alpha \): \[ 2\alpha = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z} \] \[ \alpha = \frac{3\pi}{8} + \frac{n\pi}{2} \]

Step 4: The smallest positive value of \( \alpha \) is: \[ \alpha = \frac{3\pi}{8} \]

Final Answer: The smallest positive value of \( \alpha \) is \( \frac{3\pi}{8} \).