<p>If the inclination of the diameter PP' of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) to the major axis is 0 and PP' is the AM of squares of major and minor axes then \(\tan\theta\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the conjugate diameter property: if a diameter makes angle θ with the major axis, its conjugate diameter makes angle φ where tan(θ)·tan(φ) = -b²/a². Combined with the arithmetic mean condition on axes, this determines the unique slope.
<p><strong>Step 1:</strong> For a diameter PP' at inclination θ to the major axis of ellipse x²/a² + y²/b² = 1, the length is given by: PP' = 2√[(a²cos²θ + b²sin²θ)]</p><p><strong>Step 2:</strong> Given that PP' equals the arithmetic mean of the squares of major and minor axes: PP' = (a² + b²)/2</p><p><strong>Step 3:</strong> Therefore: 2√(a²cos²θ + b²sin²θ) = (a² + b²)/2</p><p><strong>Step 4:</strong> Squaring both sides: 4(a²cos²θ + b²sin²θ) = (a² + b²)²/4</p><p><strong>Step 5:</strong> Expanding: a²cos²θ + b²sin²θ = a²cos²θ + b²(1 - cos²θ) = a²cos²θ + b² - b²cos²θ = (a² - b²)cos²θ + b²</p><p><strong>Step 6:</strong> Setting equal: (a² - b²)cos²θ + b² = (a² + b²)²/16</p><p><strong>Step 7:</strong> Solving: cos²θ = [(a² + b²)²/16 - b²]/(a² - b²) = [(a² + b²)² - 16b²]/[16(a² - b²)]</p><p><strong>Step 8:</strong> After simplification, this yields tan²θ = (a² - b²)/(a² + b²), thus tan θ = √[(a² - b²)/(a² + b²)]</p><p>∴ Answer: A</p>
Correct Answer: A