<p><strong>Paragraph for Question nos. 652 and 653</strong><br>Let \(A\) be a variable point on locus of feet of perpendicular drawn from focus upon any tangent to the curve \(|z-2|+|z+2|=6\) and \(B\) be a variable point on \((1-i)z + (1+i)\bar{z} = 10\sqrt{2}\), then<br><br>Minimum value of \(AB\) is:</p>
Step-by-Step Solution
Key Concept: The locus of feet of perpendiculars from a focus to any tangent of an ellipse is the auxiliary circle (radius = semi-major axis). Find this circle, then minimize distance from points on this circle to the given line.
<p><strong>Step 1: Identify the ellipse</strong><br>From |z-2| + |z+2| = 6, the sum of distances from foci F₁(-2,0) and F₂(2,0) is 6.<br>So 2a = 6 ⟹ a = 3, and 2c = 4 ⟹ c = 2<br>Thus b² = a² - c² = 9 - 4 = 5<br>Ellipse: x²/9 + y²/5 = 1</p><p><strong>Step 2: Find locus of feet of perpendiculars from focus to tangent</strong><br>The locus of feet of perpendiculars from any focus to any tangent of an ellipse is the auxiliary circle with radius = a.<br>Circle (locus A): x² + y² = 9</p><p><strong>Step 3: Convert the line equation for B</strong><br>(1-i)z + (1+i)z̄ = 10√2<br>Let z = x + iy:<br>(1-i)(x+iy) + (1+i)(x-iy) = 10√2<br>x + iy - ix + y + x - iy + ix + y = 10√2<br>2x + 2y = 10√2<br>x + y = 5√2</p><p><strong>Step 4: Find minimum distance AB</strong><br>Minimum distance from circle x² + y² = 9 (center O(0,0), radius 3) to line x + y = 5√2:<br>Distance from O to line: d = |0 + 0 - 5√2|/√(1² + 1²) = 5√2/√2 = 5<br>Minimum AB = d - radius = 5 - 3 = 2</p><p>∴ Answer: B</p>
Correct Answer: B