Parabola
Normal to parabola
Grade 11
Question:
<p>The normal at the point \((bt_1^2, 2bt_1)\) on a parabola meets the parabola again in the point \((bt_2^2, 2bt_2)\), then</p>
<p>\(t_2 = -t_1 - \dfrac{2}{t_1}\)</p>
<p>\(t_2 = -t_1 + \dfrac{2}{t_1}\)</p>
<p>\(t_2 = t_1 - \dfrac{2}{t_1}\)</p>
<p>\(t_2 = t_1 + \dfrac{2}{t_1}\)</p>
Step-by-Step Solution
Key Concept: The normal at a point on parabola y² = 4bx has slope -t₁ (negative reciprocal of tangent slope). Using the condition that this normal passes through another point on the parabola, we establish a relationship between t₁ and t₂ through the normal equation.
<p><strong>Step 1:</strong> For parabola y² = 4bx, the point (bt₁², 2bt₁) lies on it. The tangent at this point has slope 1/t₁, so the normal has slope <strong>-t₁</strong>.</p><p><strong>Step 2:</strong> Equation of normal at (bt₁², 2bt₁):<br/>y - 2bt₁ = -t₁(x - bt₁²)<br/>y = -t₁x + bt₁³ + 2bt₁<br/>y = -t₁x + bt₁(t₁² + 2)</p><p><strong>Step 3:</strong> The point (bt₂², 2bt₂) also lies on this normal:<br/>2bt₂ = -t₁(bt₂²) + bt₁(t₁² + 2)<br/>2t₂ = -t₁t₂² + t₁(t₁² + 2)<br/>2t₂ = -t₁t₂² + t₁³ + 2t₁</p><p><strong>Step 4:</strong> Rearranging:<br/>t₁t₂² + 2t₂ - 2t₁ - t₁³ = 0<br/>t₁t₂² - t₁³ + 2(t₂ - t₁) = 0<br/>t₁(t₂² - t₁²) + 2(t₂ - t₁) = 0<br/>t₁(t₂ - t₁)(t₂ + t₁) + 2(t₂ - t₁) = 0<br/>(t₂ - t₁)[t₁(t₂ + t₁) + 2] = 0</p><p><strong>Step 5:</strong> Since t₂ ≠ t₁ (different points):<br/><strong>t₁t₂ + t₁² + 2 = 0</strong> or equivalently <strong>t₁t₂ = -(t₁² + 2)</strong><br/>Or: <strong>t₁ + t₂ = -2/t₁</strong> and <strong>t₁t₂ = -2</strong></p><p>∴ Answer: A</p>
Correct Answer: A