Applications of Derivatives
Tangent and Normal to a Curve
Grade 12
Question:
<p>Given curve is \(x^2 + 2xy - 3y^2 = 0\). The normal to the curve at \((1, 1)\) meets the curve again at which point?</p>
<p>(3, 3)</p>
<p>(3, -1)</p>
<p>\(\left(1, -\dfrac{1}{3}\right)\)</p>
<p>All of these</p>
Step-by-Step Solution
Key Concept: Find the slope of the tangent using implicit differentiation, then use the normal's equation to find where it intersects the curve again by solving the resulting system.
<p><strong>Step 1:</strong> Find dy/dx using implicit differentiation on x² + 2xy - 3y² = 0</p><p>2x + 2y + 2x(dy/dx) - 6y(dy/dx) = 0</p><p>2x + 2y + (2x - 6y)(dy/dx) = 0</p><p>(dy/dx) = -(2x + 2y)/(2x - 6y) = -(x + y)/(x - 3y)</p><p><strong>Step 2:</strong> At point (1, 1): slope of tangent = -(1 + 1)/(1 - 3) = -2/(-2) = 1</p><p>Slope of normal = -1</p><p><strong>Step 3:</strong> Equation of normal at (1, 1): y - 1 = -1(x - 1)</p><p>y = -x + 2, or x + y = 2</p><p><strong>Step 4:</strong> Substitute y = 2 - x into curve equation x² + 2xy - 3y² = 0</p><p>x² + 2x(2 - x) - 3(2 - x)² = 0</p><p>x² + 4x - 2x² - 3(4 - 4x + x²) = 0</p><p>x² + 4x - 2x² - 12 + 12x - 3x² = 0</p><p>-4x² + 16x - 12 = 0</p><p>x² - 4x + 3 = 0</p><p>(x - 1)(x - 3) = 0</p><p>x = 1 (given point) or x = 3</p><p><strong>Step 5:</strong> When x = 3: y = 2 - 3 = -1</p><p>Verify: 9 + 2(3)(-1) - 3(1) = 9 - 6 - 3 = 0 ✓</p><p>∴ Answer: (3, -1)</p>
Correct Answer: D