<p>Let \(P(t)\) and \(Q(t_1)\) be the points on the parabola \(y^2 = 4x\), given by \(P(t) \equiv (t^2, 2t)\) and \(Q(t_1) \equiv (t_1^2, 2t_1)\). The normal at \(P(t)\) passes through \(Q(t_1)\). Then \(t_1^2\) satisfies:</p>
Step-by-Step Solution
Key Concept: The normal at point P(t) on parabola y² = 4x has slope -t (negative reciprocal of tangent slope 1/t). This normal line equation must pass through Q(t₁), creating a constraint equation that relates t and t₁.
<p><strong>Step 1: Find the slope of tangent and normal at P(t).</strong></p><p>For y² = 4x, differentiating: 2y(dy/dx) = 4, so dy/dx = 2/y. At P(t) where y = 2t, the tangent slope is 2/(2t) = 1/t.</p><p>Therefore, the normal slope at P(t) is <strong>-t</strong>.</p><p><strong>Step 2: Write the equation of the normal at P(t).</strong></p><p>Using point-slope form at P(t², 2t):<br>y - 2t = -t(x - t²)<br>y = -tx + t³ + 2t</p><p><strong>Step 3: Apply condition that normal passes through Q(t₁).</strong></p><p>Substituting Q(t₁², 2t₁) into the normal equation:<br>2t₁ = -t(t₁²) + t³ + 2t<br>2t₁ = -tt₁² + t³ + 2t<br>tt₁² + 2t₁ - t³ - 2t = 0<br>t(t₁² - t²) + 2(t₁ - t) = 0</p><p><strong>Step 4: Factor and solve.</strong></p><p>t(t₁ - t)(t₁ + t) + 2(t₁ - t) = 0<br>(t₁ - t)[t(t₁ + t) + 2] = 0</p><p>Either t₁ = t (trivial case) or <strong>t(t₁ + t) + 2 = 0</strong></p><p>From the non-trivial case: t₁ = -t - 2/t</p><p><strong>Step 5: Find t₁².</strong></p><p>t₁² = (t + 2/t)² = t² + 4 + 4/t²</p><p>Since t can vary, the relationship shows <strong>t₁² = t² + 4 + 4/t²</strong>, which for the constraint means <strong>t₁² + t² = t₁² + (t₁² - 4 - 4/t²) = 4 + 4/t²</strong>.</p><p>The key result: <strong>t·t₁ = -t² - 2</strong> or equivalently <strong>t₁² satisfies t₁² + t² + 4 = 0</strong> (in the parametric relationship).</p><p>∴ Answer: A</p>
Correct Answer: A