Question

A rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, then its dimensions are

• A. $4\sqrt{2}$, $6\sqrt{2}$
• B. $2$, $5\sqrt{2}$
• C. $4\sqrt{2}$, $5\sqrt{2}$
• D. $4\sqrt{2}$, $2$

Hint:

For any standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the sides of the maximum inscribed rectangle are always fixed structurally as $\sqrt{2}a$ and $\sqrt{2}b$. Knowing this shortcut allows you to calculate the answers directly: $\sqrt{2}(5) = 5\sqrt{2}$ and $\sqrt{2}(4) = 4\sqrt{2}$ within seconds!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the dimensions of the largest rectangle that can be inscribed in the ellipse x²/25 + y²/16 = 1.

Step 2: Key Formula or Approach:
Parametrize the ellipse as (a cos θ, b sin θ). An inscribed rectangle has dimensions 2a cos θ by 2b sin θ. Area A = 2ab sin 2θ is maximized when sin 2θ = 1, i.e., θ = π/4.

Step 3: Detailed Explanation:
From the equation, a² = 25 → a = 5; b² = 16 → b = 4. At θ = π/4: Length = 2a cos(π/4) = 10 × (1/√2) = 5√2. Breadth = 2b sin(π/4) = 8 × (1/√2) = 4√2. These are the dimensions for maximum area.

Step 4: Final Answer:
The dimensions are 4√2 and 5√2, option (C).