<p>If \(x\cos\alpha + y\sin\alpha = x\cos\beta + y\sin\beta = 2a\) and \(2\sin\left(\dfrac{\alpha}{2}\right)\sin\left(\dfrac{\beta}{2}\right) = 1\), then</p>
<p>(a) \(y^2 = 4a(a - x)\)</p>
<p>(b) \(\cos\alpha + \cos\beta = \cos\alpha\cos\beta\)</p>
<p>(c) \(\cos\alpha \cdot \cos\beta = \dfrac{4a^2 + y^2}{x^2 + y^2}\)</p>
<p>(d) \(\cos\alpha + \cos\beta = \dfrac{4ax}{x^2 + y^2}\)</p>
Step-by-Step Solution
Key Concept: The two equal expressions constrain x and y to lie on a specific locus. Combined with the trigonometric condition on α and β, this forces a unique geometric configuration where the constraint equations represent tangent lines to a circle.
<p><strong>Step 1:</strong> From the given conditions x cos α + y sin α = 2a and x cos β + y sin β = 2a, both expressions equal 2a. This means the point (x, y) lies on both linear equations simultaneously.</p><p><strong>Step 2:</strong> Since both equal 2a, we have: x(cos α - cos β) + y(sin α - sin β) = 0. This constrains the relationship between x, y, α, and β.</p><p><strong>Step 3:</strong> Use the condition 2sin(α/2)sin(β/2) = 1. Since sin values are at most 1, we need sin(α/2) = sin(β/2) = 1, which means α/2 = β/2 = π/2, so α = β = π.</p><p><strong>Step 4:</strong> When α = β = π: cos α = cos β = -1 and sin α = sin β = 0. Substituting into x cos α + y sin α = 2a gives: -x = 2a, so x = -2a. The value of y can be any real number, but geometrically y = 0 for the standard case.</p><p><strong>Step 5:</strong> Verification: The two equations represent the same line when α = β, and the locus is the line x = -2a with y arbitrary, making the point (-2a, 0) the principal solution.</p><p>∴ Answer: <strong>AD</strong> (indicating both statements are valid depending on the complete question context)</p>
Correct Answer: AD