<p>Let \(P_0\) is the parabola \(y^2 = 4x\) with vertex \(K(0,0)\), \(A\) and \(B\) are points on \(P_0\) where tangents drawn intersect at right angles. Let \(C\) be the centroid of \(\triangle ABK\). The locus of \(C\) is another parabola \(P_1\). Now the process is repeated with \(P_1\) then \(P_2, P_3, \ldots\) etc. Then the length of latus rectum of \(P_{10}\) can be expressed as \(\dfrac{a}{b}\) where \(a, b\) are co-prime natural numbers. Find the value of \((a + \log_3 b)\).</p>
Step-by-Step Solution
Key Concept: When tangents to a parabola intersect at right angles, the locus of their intersection point lies on the directrix. The centroid of triangle ABK generates a new parabola, and there's a recursive scaling pattern in the latus rectum with each iteration.
<p><strong>Step 1: Find the locus C for parabola P₀: y² = 4x</strong></p><p>For parabola y² = 4x, parametric points are A(t₁², 2t₁) and B(t₂², 2t₂).</p><p>Tangent at A: t₁y = x + t₁². Tangent at B: t₂y = x + t₂².</p><p>These tangents intersect when: t₁t₂ = -1 (condition for perpendicular tangents).</p><p><strong>Step 2: Find centroid C of triangle ABK</strong></p><p>With K(0,0), A(t₁², 2t₁), B(t₂², 2t₂):</p><p>C = ((t₁² + t₂²)/3, (2t₁ + 2t₂)/3)</p><p>Let C = (h, k). Then h = (t₁² + t₂²)/3 and k = 2(t₁ + t₂)/3.</p><p><strong>Step 3: Eliminate parameters using t₁t₂ = -1</strong></p><p>From k = 2(t₁ + t₂)/3: t₁ + t₂ = 3k/2</p><p>(t₁ + t₂)² = t₁² + t₂² + 2t₁t₂ = t₁² + t₂² - 2</p><p>9k²/4 = 3h - 2, so 3h = 9k²/4 + 2</p><p>Therefore: k² = (4/3)(h - 2/3) = (4/3)h - 8/9</p><p>Rewriting: P₁ has equation y² = (4/3)(x - 2/3), so latus rectum L₁ = 4/3.</p><p><strong>Step 4: Find the pattern for latus rectum</strong></p><p>P₀: y² = 4x, latus rectum L₀ = 4</p><p>P₁: y² = (4/3)(x - 2/3), latus rectum L₁ = 4/3</p><p>L₁/L₀ = (4/3)/4 = 1/3</p><p>Each iteration scales the latus rectum by factor 1/3.</p><p><strong>Step 5: Calculate L₁₀</strong></p><p>Lₙ = 4 × (1/3)ⁿ</p><p>L₁₀ = 4 × (1/3)¹⁰ = 4/3¹⁰</p><p><strong>Step 6: Express in form a/b where gcd(a,b) = 1</strong></p><p>L₁₀ = 4/3¹⁰ = 4/59049</p><p>gcd(4, 59049) = 1 (since 59049 = 3¹⁰ is odd)</p><p>Therefore a = 4, b = 59049 = 3¹⁰</p><p><strong>Step 7: Calculate final answer</strong></p><p>a + log₃(b) = 4 + log₃(3¹⁰) = 4 + 10 = 14</p><p>∴ Answer: 14</p>
Correct Answer: 14