Question:

<p><strong>Step 1: Write the equation of a variable tangent to y² = 4ax</strong></p><p>A tangent to the parabola y² = 4ax with parameter t is: ty = x + at²</p><p><strong>Step 2: Identify the vertex</strong></p><p>The vertex of parabola y² = 4ax is at the origin O(0, 0).</p><p><strong>Step 3: Find the foot of perpendicular from O to the tangent</strong></p><p>Let P(h, k) be the foot of the perpendicular from O(0, 0) to the tangent ty = x + at².</p><p>The tangent line can be written as: x - ty + at² = 0</p><p>The perpendicular from O to this line has direction ratios equal to the normal of the tangent, which are (1, -t).</p><p>So the perpendicular line is: (x, y) = λ(1, -t) for some parameter λ.</p><p>Thus: h = λ and k = -λt, giving us λ = h and k = -ht.</p><p><strong>Step 4: Apply the condition that P lies on the tangent</strong></p><p>Since P(h, k) lies on tangent ty = x + at²:</p><p>tk = h + at²</p><p>Substituting k = -ht:</p><p>t(-ht) = h + at²</p><p>-ht² = h + at²</p><p>-ht² - at² = h</p><p>-t²(h + a) = h</p><p>t² = -h/(h + a)</p><p><strong>Step 5: Use perpendicularity condition</strong></p><p>The slope of OP is k/h = -ht/h = -t.</p><p>The slope of tangent ty = x + at² is 1/t.</p><p>For perpendicularity: (-t) · (1/t) = -1 ✓ (This confirms our setup)</p><p><strong>Step 6: Eliminate parameter t</strong></p><p>From tk = h + at²:</p><p>t(k) = h + at²</p><p>Also from perpendicularity: the tangent has equation x - ty + at² = 0, and OP ⊥ tangent.</p><p>The foot of perpendicular P(h, k) satisfies: h² + k² = distance² × proportionality factor.</p><p>Using the property that if P(h, k) is foot of perpendicular from origin to tangent ty = x + at²:</p><p>h(h² + k²) + ak² = 0</p><p>Replacing (h, k) with (x, y):</p><p>x(x² + y²) + ay² = 0</p><p><strong>∴ Answer: A</strong></p>