<p>Consider an ellipse, whose centre is at the origin and its major axis is along the \(x\)-axis. If its eccentricity is \(3/5\) and the distance between its foci is 6, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is</p>
Step-by-Step Solution
Key Concept: For an ellipse with semi-major axis a and semi-minor axis b, the distance between foci is 2c where c = ae. Using e = 3/5 and 2c = 6, find a and b, then calculate the area of the quadrilateral formed by the four vertices (endpoints of major and minor axes).
<p><strong>Step 1:</strong> Set up the relationship. For an ellipse with semi-major axis <em>a</em>, semi-minor axis <em>b</em>, and eccentricity <em>e</em>: e = c/a, where c² = a² − b².</p><p><strong>Step 2:</strong> Use given data. Distance between foci = 2c = 6, so c = 3. Also, e = 3/5.</p><p><strong>Step 3:</strong> Find <em>a</em>. From e = c/a: (3/5) = 3/a, therefore a = 5.</p><p><strong>Step 4:</strong> Find <em>b</em>. From c² = a² − b²: 9 = 25 − b², so b² = 16, thus b = 4.</p><p><strong>Step 5:</strong> Identify the quadrilateral. The vertices of the ellipse are at (±5, 0) and (0, ±4). These four points form a rectangle with length 2a = 10 and width 2b = 8.</p><p><strong>Step 6:</strong> Calculate area. Area = (2a)(2b) = 10 × 8 = 80 sq. units.</p><p>∴ Answer: D (80)</p>
Correct Answer: D