<p>If \(a \neq 0\) and the line \(2bx + 3cy + 4d = 0\) passes through the points of intersection of the parabolas \(y^2 = 4ax\) and \(x^2 = 4ay\), then</p>
<p>\(d^2 + (2b + 3c)^2 = 0\)</p>
<p>\(d^2 + (3b + 2c)^2 = 0\)</p>
<p>\(d^2 + (2b - 3c)^2 = 0\)</p>
<p>\(d^2 + (3b - 2c)^2 = 0\)</p>
Step-by-Step Solution
Key Concept: The line through intersection points of two conics must satisfy the condition that it represents the 'chord of intersection'. For parabolas y² = 4ax and x² = 4ay, find their intersection points first, then use collinearity to establish a relationship between b, c, and d.
<p><strong>Step 1:</strong> Find intersection points of y² = 4ax and x² = 4ay.</p><p>From y² = 4ax, substitute into x² = 4ay:</p><p>x² = 4a√(4ax) is complex, so use parametric approach.</p><p>From y² = 4ax: points are (at², 2at)</p><p>From x² = 4ay: points are (2au, 4au²/a) = (2au, 4au)</p><p><strong>Step 2:</strong> Solve simultaneously. If (x,y) satisfies both:</p><p>y² = 4ax and x² = 4ay</p><p>Then y²/x² = 4ax/4ay = x/y, so y³ = x³, giving y = x (for real intersection)</p><p>Also, the origin (0,0) satisfies both.</p><p>Substituting y = x into y² = 4ax: x² = 4ax ⟹ x(x - 4a) = 0</p><p>So intersection points are: (0,0) and (4a, 4a)</p><p><strong>Step 3:</strong> The line 2bx + 3cy + 4d = 0 passes through both (0,0) and (4a, 4a).</p><p>At (0,0): 4d = 0 ⟹ d = 0</p><p>At (4a, 4a): 2b(4a) + 3c(4a) + 0 = 0</p><p>⟹ 8ab + 12ac = 0</p><p>⟹ 2b + 3c = 0 (dividing by 4a, since a ≠ 0)</p><p><strong>∴ Answer: A (The relationship is 2b + 3c = 0 and d = 0)</strong></p>
Correct Answer: A