<p>The equation of ellipse is \(3x^2 + 5y^2 = 32\). Find the area of triangle \(\triangle ABC\) formed by the tangent and normal to the ellipse at point \((2, 2)\) with the axes (in sq. units).</p>
Step-by-Step Solution
Key Concept: Find the tangent and normal lines at (2,2) on the ellipse, then calculate the area of the triangle formed by these two lines and their intercepts with the coordinate axes using the formula for triangle area with known vertices.
<p><strong>Step 1: Verify point on ellipse</strong></p><p>Check: 3(2)² + 5(2)² = 12 + 20 = 32 ✓</p><p><strong>Step 2: Find equation of tangent at (2,2)</strong></p><p>For ellipse 3x² + 5y² = 32, tangent at (x₀,y₀) is: 3xx₀ + 5yy₀ = 32</p><p>At (2,2): 3x(2) + 5y(2) = 32 → <strong>3x + 5y = 16</strong></p><p><strong>Step 3: Find equation of normal at (2,2)</strong></p><p>Normal is perpendicular to tangent. Slope of tangent = -3/5, so slope of normal = 5/3</p><p>Normal: y - 2 = (5/3)(x - 2) → <strong>5x - 3y - 4 = 0</strong></p><p><strong>Step 4: Find vertices of triangle</strong></p><p>• Point A: Tangent meets x-axis (y=0): 3x = 16 → x = 16/3, so A = (16/3, 0)</p><p>• Point B: Normal meets x-axis (y=0): 5x - 4 = 0 → x = 4/5, so B = (4/5, 0)</p><p>• Point C: Intersection of tangent and normal: Solving 3x + 5y = 16 and 5x - 3y = 4</p><p>Multiply first by 3: 9x + 15y = 48; multiply second by 5: 25x - 15y = 20</p><p>Adding: 34x = 68 → x = 2, y = 2, so C = (2, 2)</p><p><strong>Step 5: Calculate area using base-height formula</strong></p><p>Base AB = |16/3 - 4/5| = |80/15 - 12/15| = 68/15</p><p>Height = perpendicular distance from C(2,2) to x-axis = 2</p><p>Area = (1/2) × base × height = (1/2) × (68/15) × 2 = <strong>68/15 ≈ 4.533 sq. units</strong></p>
Correct Answer: 4.533