<p>A square has one vertex at the vertex of the parabola \(y^2 = 4ax\) and the diagonal through the vertex lies along the axis of the parabola. If the ends of the other diagonal lie on the parabola, the coordinates of the vertices of the square are</p>
Step-by-Step Solution
Key Concept: Place the square with one vertex at origin (parabola's vertex) and use the diagonal constraint: if a diagonal lies along the x-axis, the square has sides at 45° to axes. The symmetry of the parabola about the x-axis ensures both ends of the other diagonal lie on the parabola simultaneously.
<p><strong>Step 1:</strong> Let the square have vertex at origin O(0,0). Since one diagonal lies along the x-axis (axis of parabola), let the other two adjacent vertices be at equal distances along lines at ±45° to the x-axis.</p><p><strong>Step 2:</strong> For a square with side length s, if one vertex is at origin and one diagonal lies along x-axis, the four vertices are: O(0,0), A(d,0), and two vertices B, C at equal heights ±h on a line perpendicular to OA, where the diagonal through B and C is perpendicular to OA.</p><p><strong>Step 3:</strong> By symmetry about x-axis, let B = (x₀, y₀) and C = (x₀, -y₀) be on the parabola. Then: y₀² = 4ax₀. For a square, if diagonal OA has length d along x-axis, then d = 2x₀ and the perpendicular diagonal has equal length, so 2y₀ = 2x₀, giving y₀ = x₀.</p><p><strong>Step 4:</strong> Substituting y₀ = x₀ into parabola equation: x₀² = 4ax₀, so x₀ = 4a (taking x₀ ≠ 0). Thus y₀ = 4a.</p><p><strong>Step 5:</strong> The four vertices of the square are:</p><p>• <strong>O(0, 0)</strong> - vertex at parabola's vertex</p><p>• <strong>A(8a, 0)</strong> - opposite vertex on x-axis</p><p>• <strong>B(4a, 4a)</strong> - on parabola</p><p>• <strong>C(4a, -4a)</strong> - on parabola</p><p>∴ Answer: (0,0), (8a,0), (4a,4a), (4a,-4a)</p>
Correct Answer: A,B,C,D