<p>If the normal to the ellipse \(3x^2 + 4y^2 = 12\) at a point P on it is parallel to the line, \(2x + y = 4\) and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to __________ (up to two decimal places).</p>
Step-by-Step Solution
Key Concept: The normal at point P has slope -1/2 (parallel to given line), so the tangent has slope 2. Use the tangent equation passing through Q(4,4) to find point P on the ellipse, then calculate distance PQ.
<p><strong>Step 1:</strong> Convert ellipse to standard form: x²/4 + y²/3 = 1</p><p><strong>Step 2:</strong> The normal is parallel to 2x + y = 4, so normal slope = -2. Therefore, tangent slope = 1/2 (negative reciprocal).</p><p><strong>Step 3:</strong> For ellipse x²/4 + y²/3 = 1, the tangent at point P(x₀, y₀) is: (xx₀)/4 + (yy₀)/3 = 1</p><p><strong>Step 4:</strong> This tangent passes through Q(4,4): (4x₀)/4 + (4y₀)/3 = 1, giving: x₀ + (4y₀)/3 = 1 ... (i)</p><p><strong>Step 5:</strong> Tangent slope = -3x₀/(4y₀) = 1/2, so: -6x₀ = 4y₀, giving: y₀ = -3x₀/2 ... (ii)</p><p><strong>Step 6:</strong> Substitute (ii) into (i): x₀ + 4(-3x₀/2)/3 = 1 → x₀ - 2x₀ = 1 → x₀ = -1</p><p><strong>Step 7:</strong> From (ii): y₀ = -3(-1)/2 = 3/2. So P(-1, 3/2)</p><p><strong>Step 8:</strong> Verify P is on ellipse: (-1)²/4 + (3/2)²/3 = 1/4 + 9/12 = 1/4 + 3/4 = 1 ✓</p><p><strong>Step 9:</strong> Distance PQ = √[(4-(-1))² + (4-3/2)²] = √[25 + (5/2)²] = √[25 + 25/4] = √[125/4] = 5√5/2 ≈ 5.59</p><p>∴ Answer: <strong>5.59</strong></p>
Correct Answer: 5