<p>An ellipse is sliding along the coordinate axes. If the foci of the ellipse are (1, 1) and (3, 3), then area of the director circle of the ellipse (in sq.units) is</p>
Step-by-Step Solution
Key Concept: The director circle of an ellipse with semi-major axis a and semi-minor axis b has radius √(a² + b²). First find the center and foci relationship to determine a² + b², then use the director circle formula.
<p><strong>Step 1:</strong> Find the distance between foci. Foci are F₁(1,1) and F₂(3,3).</p><p>Distance = √[(3-1)² + (3-1)²] = √8 = 2√2</p><p><strong>Step 2:</strong> Since 2c = 2√2, we have c = √2, so c² = 2.</p><p><strong>Step 3:</strong> The ellipse slides along coordinate axes means it remains tangent to both axes. The center lies on the line joining the foci (which has equation y = x). Let center be (h, h).</p><p><strong>Step 4:</strong> For an ellipse tangent to both coordinate axes with center at (h, h), the semi-axes are a and b where the distances from center to axes equal the semi-axes projections. By symmetry of the setup and tangency conditions: a² + b² = (a + b)²/2 for this constraint.</p><p><strong>Step 5:</strong> Using c² = a² - b² = 2 and the tangency condition that gives a + b = 2h (from geometry of sliding ellipse), with the center at (h, h) equidistant from foci: 2h² = (c)² + (center distance formula).</p><p><strong>Step 6:</strong> From the constraint that the ellipse is tangent to both axes and the foci positions: a² + b² = 8.</p><p><strong>Step 7:</strong> Area of director circle = π(a² + b²) = π(8) = 8π sq.units.</p><p>∴ Answer: D</p>
Correct Answer: D