<p>If the tangent to the ellipse \(x^2 + 4y^2 = 16\) at the point \(P\phi\) is a normal to the circle \(x^2 + y^2 - 8x - 4y = 0\), then \(\phi\) is equal to</p>
Step-by-Step Solution
Key Concept: A line is both a tangent to the ellipse and a normal to the circle simultaneously. Use the tangent equation at parameter φ on the ellipse, then verify it passes through the circle's center (making it a normal).
<p><strong>Step 1:</strong> Rewrite ellipse in standard form: x²/16 + y²/4 = 1. Point P at parameter φ is P(4cosφ, 2sinφ).</p><p><strong>Step 2:</strong> Tangent at P(4cosφ, 2sinφ) to the ellipse is: (xcosφ)/4 + (ysinφ)/2 = 1, or xcosφ + 2ysinφ = 4.</p><p><strong>Step 3:</strong> Rewrite circle: x² + y² - 8x - 4y = 0 → (x-4)² + (y-2)² = 20. Center is C(4, 2).</p><p><strong>Step 4:</strong> For the tangent to be a normal to the circle, it must pass through center C(4, 2). Substitute into tangent equation:</p><p>4cosφ + 2(2)sinφ = 4</p><p>4cosφ + 4sinφ = 4</p><p>cosφ + sinφ = 1</p><p><strong>Step 5:</strong> Square both sides: cos²φ + 2cosφsinφ + sin²φ = 1 → 1 + 2cosφsinφ = 1 → sin(2φ) = 0.</p><p>This gives 2φ = 0°, 180°, 360°... so φ = 0°, 90°, 180°...</p><p><strong>Step 6:</strong> Check φ = 90°: cosφ + sinφ = 0 + 1 = 1 ✓</p><p><strong>Step 7:</strong> Check φ = 0°: cosφ + sinφ = 1 + 0 = 1 ✓</p><p>∴ Answer: φ = π/2 (or 90°) [or φ = 0 depending on options given]</p>
Correct Answer: A