Parabola
Locus of Midpoint of Normal Chords
Grade 11
Question:
<p>Find the locus of the middle points of the normal chords of the parabola \(y^2 = 4ax\).</p>
<p>\(x + 2a = \dfrac{y^2}{2a} + \dfrac{8a^3}{y^2}\)</p>
<p>\(2ay^2(x - 2ay) = 8a^4 + y^4\)</p>
<p>\(y^2(x+2a) = 8a^3 + y^4\)</p>
<p>\(2ay^2(x-2a) = 8a^4 - y^4\)</p>
Step-by-Step Solution
Key Concept: A normal chord has its endpoints on the parabola such that the normals at these points are collinear. Use the parametric form and normal equation to find the relationship between the two parameters, then find the midpoint locus by eliminating parameters.
<p><strong>Step 1:</strong> For parabola y² = 4ax, parametric form: point P = (at₁², 2at₁) and Q = (at₂², 2at₂)</p><p><strong>Step 2:</strong> The slope of normal at (at², 2at) is -t. For a normal chord, the chord PQ itself must be a normal, giving the condition: t₁ + t₂ + t₁t₂ = 0</p><p><strong>Step 3:</strong> Midpoint M of chord PQ: h = a(t₁² + t₂²)/2 and k = a(t₁ + t₂)</p><p><strong>Step 4:</strong> From t₁ + t₂ + t₁t₂ = 0, we get t₁t₂ = -(t₁ + t₂) = -k/a</p><p><strong>Step 5:</strong> Therefore: (t₁ + t₂)² = (k/a)² and t₁² + t₂² = (t₁ + t₂)² - 2t₁t₂ = k²/a² + 2k/a</p><p><strong>Step 6:</strong> Substituting into h: h = a/2 · [k²/a² + 2k/a] = k²/(2a) + k</p><p><strong>Step 7:</strong> Rearranging: y² = 2a(x - a) or <strong>y² = 2a(h - a)</strong></p><p>∴ Answer: B</p>
Correct Answer: B