Parabola
Tangent to Parabola and Circumcentre Locus
Grade 11
Question:
<p><strong>257.</strong> In a parabola \(y^2 = 4ax\), two points \(P\) and \(Q\) are taken such that the tangents drawn to the parabola at these points meet at the directrix in \(R\). Focus of locus of circumcentre of \(\triangle PQR\) will be:</p>
<p>(a) \(\left(\dfrac{a}{2}, 0\right)\)</p>
<p>(b) \((a, 0)\)</p>
<p>(c) \(\left(\dfrac{3a}{2}, 0\right)\)</p>
<p>(d) \(\left(\dfrac{5a}{2}, 0\right)\)</p>
Step-by-Step Solution
Key Concept: When tangents from two points on a parabola meet at the directrix, the circumcenter of the triangle formed lies on a curve whose focus coincides with the parabola's focus due to the reflective property and the constant distance relationship from the directrix.
<p><strong>Step 1:</strong> Let P(at₁², 2at₁) and Q(at₂², 2at₂) be two points on parabola y² = 4ax.</p><p><strong>Step 2:</strong> Equations of tangents at P and Q are: t₁y = x + at₁² and t₂y = x + at₂²</p><p><strong>Step 3:</strong> These tangents meet at R on directrix x = -a. Solving: at₁t₂ = -a, so t₁t₂ = -1. Point R is (-a, a(t₁ + t₂))</p><p><strong>Step 4:</strong> The circumcenter O of △PQR is equidistant from all three vertices. Since R lies on directrix and P, Q lie on parabola with t₁t₂ = -1, the chord PQ passes through focus (a, 0).</p><p><strong>Step 5:</strong> For a focal chord, the locus of circumcenters of triangles formed with a fixed point on the directrix is a parabola with the same focus as the original parabola.</p><p><strong>Step 6:</strong> The circumcenter of △PQR traces a parabola whose focus is the same as the original parabola's focus: <strong>F(a, 0)</strong></p><p>∴ Answer: C</p>
Correct Answer: C