Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>The normal to the curve, \(x^2 + 2xy - 3y^2 = 0\) at \((1, 1)\)</p>
<p>meets the curve again in the second quadrant.</p>
<p>meets the curve again in the third quadrant.</p>
<p>meets the curve again in the fourth quadrant.</p>
<p>does not meet the curve again.</p>
Step-by-Step Solution
Key Concept: Find the slope of the tangent using implicit differentiation, then use the fact that the normal is perpendicular to the tangent (slope_normal = -1/slope_tangent).
<p><strong>Step 1:</strong> Differentiate implicitly: x² + 2xy - 3y² = 0</p><p>2x + 2(y + x·dy/dx) - 6y·dy/dx = 0</p><p>2x + 2y + 2x·dy/dx - 6y·dy/dx = 0</p><p><strong>Step 2:</strong> Solve for dy/dx:</p><p>2x·dy/dx - 6y·dy/dx = -2x - 2y</p><p>dy/dx(2x - 6y) = -2(x + y)</p><p>dy/dx = -2(x + y)/(2x - 6y) = -(x + y)/(x - 3y)</p><p><strong>Step 3:</strong> At point (1, 1):</p><p>dy/dx = -(1 + 1)/(1 - 3) = -2/(-2) = 1</p><p><strong>Step 4:</strong> Slope of normal = -1/slope of tangent = -1/1 = -1</p><p><strong>Step 5:</strong> Equation of normal at (1, 1) with slope -1:</p><p>y - 1 = -1(x - 1)</p><p>y - 1 = -x + 1</p><p><strong>∴ Answer: x + y = 2 (Option C)</strong></p>
Correct Answer: C