<p>The tangent and normal to the ellipse \(3x^2 + 5y^2 = 32\) at the point P(2, 2) meet the \(x\)-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is __________ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: Find the equations of tangent and normal at point P(2,2) on the ellipse, determine where they intersect the x-axis to get points Q and R, then calculate the area of triangle PQR using the base QR on x-axis and height equal to the y-coordinate of P.
<p><strong>Step 1: Verify P(2,2) lies on ellipse</strong></p><p>3(2)² + 5(2)² = 12 + 20 = 32 ✓</p><p><strong>Step 2: Find equation of tangent at P(2,2)</strong></p><p>For ellipse 3x² + 5y² = 32, tangent at (x₀, y₀) is: 3x·x₀ + 5y·y₀ = 32</p><p>At P(2,2): 3x(2) + 5y(2) = 32 → 6x + 10y = 32 → 3x + 5y = 16</p><p><strong>Step 3: Find point Q (tangent meets x-axis)</strong></p><p>Set y = 0: 3x = 16 → x = 16/3</p><p>So Q = (16/3, 0)</p><p><strong>Step 4: Find equation of normal at P(2,2)</strong></p><p>Slope of tangent = -3/5, so slope of normal = 5/3</p><p>Normal equation: y - 2 = (5/3)(x - 2) → 5x - 3y = 4</p><p><strong>Step 5: Find point R (normal meets x-axis)</strong></p><p>Set y = 0: 5x = 4 → x = 4/5</p><p>So R = (4/5, 0)</p><p><strong>Step 6: Calculate area of triangle PQR</strong></p><p>Base QR = |16/3 - 4/5| = |80/15 - 12/15| = 68/15</p><p>Height = y-coordinate of P = 2</p><p>Area = (1/2) × (68/15) × 2 = 68/15 = 4.533</p><p>∴ Answer: <strong>4.533</strong></p>
Correct Answer: 4