Parabola
Tangent to Parabola
Grade 11
Question:
<p>Given the tangents to the curve <br/> \(y = (x-2)^2 - 1\) at its points of intersection with the line \(x - y = 3\). Find the point of intersection of tangents (i.e., the point C).</p>
<p>\(\left(\dfrac{5}{2}, -1\right)\)</p>
<p>\(\left(\dfrac{3}{2}, -1\right)\)</p>
<p>\(\left(\dfrac{5}{2}, 1\right)\)</p>
<p>\(\left(\dfrac{3}{2}, 1\right)\)</p>
Step-by-Step Solution
Key Concept: Find intersection points of the parabola with the line, then find tangent equations at those points and solve simultaneously to get their intersection—this point C is independent of the specific coordinates and lies on the directrix-related locus of the parabola.
<p><strong>Step 1:</strong> Find intersection points of y = (x-2)² - 1 and x - y = 3.</p><p>Substitute y = x - 3 into the parabola: x - 3 = (x-2)² - 1</p><p>x - 3 = x² - 4x + 4 - 1</p><p>0 = x² - 5x + 6 = (x-2)(x-3)</p><p>So x = 2 or x = 3</p><p>Points: P₁(2, -1) and P₂(3, 0)</p><p><strong>Step 2:</strong> Find tangent slopes at these points.</p><p>dy/dx = 2(x-2)</p><p>At P₁(2, -1): slope m₁ = 2(2-2) = 0</p><p>At P₂(3, 0): slope m₂ = 2(3-2) = 2</p><p><strong>Step 3:</strong> Write tangent equations.</p><p>At P₁: y - (-1) = 0(x - 2) → y = -1</p><p>At P₂: y - 0 = 2(x - 3) → y = 2x - 6</p><p><strong>Step 4:</strong> Find intersection C of tangents.</p><p>-1 = 2x - 6</p><p>2x = 5 → x = 5/2</p><p>∴ Point C = <strong>(5/2, -1)</strong> or **(2.5, -1)**</p>
Correct Answer: A