Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>Let \(C\) be a curve given by \(y(x) = 1 + \sqrt{4x - 3}\), \(x > 3/4\). If \(P\) is a point on \(C\), such that the tangent at \(P\) has slope \(2/3\), then a point through which the normal at \(P\) passes, is</p>
<p>\((1, 7)\)</p>
<p>\((3, -4)\)</p>
<p>\((4, -3)\)</p>
<p>\((2, 3)\)</p>
Step-by-Step Solution
Key Concept: Find the point P where the tangent slope equals 2/3 by differentiating y(x), then use the negative reciprocal slope to write the normal line equation and identify a point it passes through.
<p><strong>Step 1:</strong> Find dy/dx for y = 1 + √(4x - 3)</p><p>dy/dx = (1/2) · (4x - 3)^(-1/2) · 4 = 2/√(4x - 3)</p><p><strong>Step 2:</strong> Set dy/dx = 2/3 to find point P</p><p>2/√(4x - 3) = 2/3</p><p>√(4x - 3) = 3</p><p>4x - 3 = 9</p><p>x = 3</p><p><strong>Step 3:</strong> Find y-coordinate at P</p><p>y = 1 + √(4(3) - 3) = 1 + √9 = 1 + 3 = 4</p><p>So P = (3, 4)</p><p><strong>Step 4:</strong> Find equation of normal at P</p><p>Slope of normal = -1/(2/3) = -3/2</p><p>Using point-slope form: y - 4 = (-3/2)(x - 3)</p><p>y = 4 - (3/2)(x - 3) = 4 - (3/2)x + 9/2</p><p>y = (17/2) - (3/2)x</p><p><strong>Step 5:</strong> Identify a point on this normal line (verify with answer choices)</p><p>When x = 1: y = 17/2 - 3/2 = 14/2 = 7, giving point (1, 7)</p><p>∴ Answer: C</p>
Correct Answer: C