<p>It is given that eccentricity is \(e = \dfrac{3}{5}\) and the distance between the foci is \(2ae = 6\). Find the area of the quadrilateral formed by the axes of the ellipse.</p>
Step-by-Step Solution
Key Concept: The axes of an ellipse are the major and minor axes which are perpendicular bisectors of each other, forming a rhombus with diagonals of length 2a and 2b. Use e = 3/5 and 2ae = 6 to find a and b, then calculate the area as (1/2)×d₁×d₂.
<p><strong>Step 1:</strong> Find the value of <em>a</em> using the given information.</p><p>Given: eccentricity e = 3/5 and distance between foci = 2ae = 6</p><p>From 2ae = 6: ae = 3</p><p>So: a(3/5) = 3 → a = 5</p><p><strong>Step 2:</strong> Find the value of <em>b</em> using the relationship b² = a²(1 - e²).</p><p>b² = a²(1 - e²) = 25(1 - 9/25) = 25(16/25) = 16</p><p>Therefore: b = 4</p><p><strong>Step 3:</strong> Identify the quadrilateral formed by the axes.</p><p>The major axis (length 2a = 10) and minor axis (length 2b = 8) are perpendicular and bisect each other at the center. The four endpoints form a rhombus with diagonals of length 2a = 10 and 2b = 8.</p><p><strong>Step 4:</strong> Calculate the area of the rhombus.</p><p>Area = (1/2) × d₁ × d₂ = (1/2) × 10 × 8 = 40</p><p>∴ Answer: C (Area = 40 square units)</p>
Correct Answer: C