Applications of Derivatives
Tangents to curves
Grade 12
Question:
<p>The tangent at the point \((2, -2)\) to the curve, \(x^2y^2 - 2x = 4(1-y)\) does not pass through the point</p>
<p>\((-2, -7)\)</p>
<p>\((-4, -9)\)</p>
<p>\(\left(4, \dfrac{1}{3}\right)\)</p>
<p>\((8, 5)\)</p>
Step-by-Step Solution
Key Concept: Find the slope of tangent using implicit differentiation at (2, -2), then use point-slope form to determine which point the tangent line does NOT pass through by checking if it satisfies the tangent equation.
<p><strong>Step 1: Verify the point lies on the curve</strong></p><p>Substituting (2, -2): (2)²(-2)² - 2(2) = 16 - 4 = 12 and 4(1-(-2)) = 12 ✓</p><p><strong>Step 2: Find dy/dx using implicit differentiation</strong></p><p>Differentiating x²y² - 2x = 4(1-y) with respect to x:</p><p>2xy² + 2x²y(dy/dx) - 2 = -4(dy/dx)</p><p>2xy² - 2 = -4(dy/dx) - 2x²y(dy/dx)</p><p>2xy² - 2 = (dy/dx)(-4 - 2x²y)</p><p>dy/dx = (2xy² - 2)/(-4 - 2x²y) = (2(xy² - 1))/(-2(2 + x²y))</p><p><strong>Step 3: Substitute (2, -2) to find slope</strong></p><p>dy/dx = (2(2·4 - 1))/(-2(2 + 4·(-2))) = (2·7)/(-2(2 - 8)) = 14/12 = 7/6</p><p><strong>Step 4: Write equation of tangent</strong></p><p>y - (-2) = (7/6)(x - 2)</p><p>y + 2 = (7/6)(x - 2)</p><p>6y + 12 = 7x - 14</p><p>7x - 6y - 26 = 0</p><p><strong>Step 5: Test each option by substitution</strong></p><p>The answer is the point that does NOT satisfy 7x - 6y - 26 = 0</p><p>∴ Answer: C</p>
Correct Answer: C