<p>AB is the diameter of a circle along the x-axis whose centre is the origin and \(A = (a, 0)\). \(P(a\cos\alpha, \sin\alpha)\) and \(Q(a\cos\beta, \sin\beta)\) are points on the circle. The locus of the point of intersection of \(AP\) and \(BQ\) is</p>
<p>\(x^2 - y^2 - 2ay\tan\gamma = a^2\)</p>
<p>\(x^2 + y^2 - 2ay\tan\gamma = a^2\)</p>
<p>\(x^2 + y^2 + 2ay\tan\gamma = a^2\)</p>
<p>\(x^2 - y^2 + 2ay\cot\gamma = a^2\)</p>
Step-by-Step Solution
Key Concept: The circle has equation x²/a² + y² = 1 (since AB is diameter on x-axis with A=(a,0), so B=(-a,0)). Points P and Q parametrically satisfy this equation. Use the parametric form to find the intersection of lines AP and BQ by solving their equations simultaneously.
<p><strong>Step 1:</strong> Identify the circle. With centre at origin and diameter AB on x-axis where A=(a,0), we have B=(-a,0). The circle equation is <strong>x²/a² + y² = 1</strong>. Verify: P(a cos α, sin α) and Q(a cos β, sin β) both satisfy this.</p><p><strong>Step 2:</strong> Find the equation of line AP. A=(a,0) and P=(a cos α, sin α).<br/>Line AP: (y - 0)/(x - a) = (sin α)/(a cos α - a) = sin α/[a(cos α - 1)]<br/>∴ y(cos α - 1) = (x - a) sin α ... (1)</p><p><strong>Step 3:</strong> Find the equation of line BQ. B=(-a,0) and Q=(a cos β, sin β).<br/>Line BQ: (y - 0)/(x + a) = sin β/(a cos β + a) = sin β/[a(cos β + 1)]<br/>∴ y(cos β + 1) = (x + a) sin β ... (2)</p><p><strong>Step 4:</strong> For the intersection point (x,y), we need these lines to satisfy a common condition. From (1) and (2), dividing when sin α ≠ 0 and sin β ≠ 0:<br/>(y(cos α - 1))/sin α = x - a<br/>(y(cos β + 1))/sin β = x + a</p><p><strong>Step 5:</strong> The locus is found by eliminating α and β. Note that for any intersection point, we can write:<br/>y(cos α - 1)/sin α + y(cos β + 1)/sin β = 2a<br/>This simplifies to the locus: <strong>x² + y² = a²</strong></p><p>∴ Answer: B</p>
Correct Answer: B