Parabola
Tangent to Parabola
Grade 11
Question:
<p>The tangents to the curve \(y = (x-2)^2 - 1\) at its points of intersection with the line \(x - y = 3\), intersect at the point</p>
<p>\(\left(\dfrac{5}{2}, 1\right)\)</p>
<p>\(\left(-\dfrac{5}{2}, -1\right)\)</p>
<p>\(\left(\dfrac{5}{2}, -1\right)\)</p>
<p>\(\left(-\dfrac{5}{2}, 1\right)\)</p>
Step-by-Step Solution
Key Concept: Find intersection points of the parabola and line, then find tangent equations at those points. The intersection of two tangent lines can be found by solving their equations simultaneously.
<p><strong>Step 1: Find intersection points of parabola and line</strong></p><p>Parabola: y = (x-2)² - 1</p><p>Line: x - y = 3, so y = x - 3</p><p>Substituting: (x-2)² - 1 = x - 3</p><p>x² - 4x + 4 - 1 = x - 3</p><p>x² - 5x + 6 = 0</p><p>(x - 2)(x - 3) = 0</p><p>x = 2 or x = 3</p><p>When x = 2: y = -1 → Point A(2, -1)</p><p>When x = 3: y = 0 → Point B(3, 0)</p></p><p><strong>Step 2: Find slopes of tangent lines</strong></p><p>For y = (x-2)² - 1: dy/dx = 2(x-2)</p><p>At x = 2: slope m₁ = 2(2-2) = 0</p><p>At x = 3: slope m₂ = 2(3-2) = 2</p></p><p><strong>Step 3: Write equations of tangents</strong></p><p>Tangent at A(2, -1) with slope 0:</p><p>y + 1 = 0(x - 2) → y = -1</p><p>Tangent at B(3, 0) with slope 2:</p><p>y - 0 = 2(x - 3) → y = 2x - 6</p></p><p><strong>Step 4: Find intersection of tangents</strong></p><p>Setting y = -1 in y = 2x - 6:</p><p>-1 = 2x - 6</p><p>2x = 5</p><p>x = 5/2</p><p>∴ Answer: (5/2, -1) or equivalent form based on option C</p>
Correct Answer: C