Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>The normal to the curve \(y(x-2)(x-3) = x + 6\) at the point where the curve intersects the y-axis passes through the point:</p>
<p>\(\left(\dfrac{1}{2}, \dfrac{1}{2}\right)\)</p>
<p>\(\left(\dfrac{1}{2}, -\dfrac{1}{3}\right)\)</p>
<p>\(\left(\dfrac{1}{2}, \dfrac{1}{3}\right)\)</p>
<p>\(\left(-\dfrac{1}{2}, -\dfrac{1}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Find the y-intercept by setting x=0, then calculate dy/dx at that point to get the normal's slope, and use point-slope form to find which point the normal passes through.
<p><strong>Step 1:</strong> Find the y-intercept. Set x = 0 in the equation y(x-2)(x-3) = x + 6:</p><p>y(0-2)(0-3) = 0 + 6</p><p>y(-2)(-3) = 6</p><p>6y = 6 → y = 1</p><p>Point of intersection with y-axis: (0, 1)</p><p><strong>Step 2:</strong> Find dy/dx using implicit differentiation on y(x-2)(x-3) = x + 6:</p><p>Expanding left side: y(x² - 5x + 6) = x + 6</p><p>Differentiating: dy/dx(x² - 5x + 6) + y(2x - 5) = 1</p><p>At (0, 1): dy/dx(6) + 1(-5) = 1</p><p>6(dy/dx) = 6 → dy/dx = 1</p><p><strong>Step 3:</strong> The slope of the normal is the negative reciprocal of the tangent slope:</p><p>Slope of normal = -1/1 = -1</p><p><strong>Step 4:</strong> Equation of normal at (0, 1) with slope -1:</p><p>y - 1 = -1(x - 0)</p><p>y = -x + 1</p><p>∴ The normal passes through points satisfying y = -x + 1 (such as (1, 0), (2, -1), etc., depending on the options given)</p>
Correct Answer: A