Consider an ellipse `x^(2)/a^(2) + y^(2)/b^(2)=1`
What is the area of the greatest rectangle that can be inscribed in the ellipse?

Options

A. ab
B. 2ab
C. ab/2
D. `sqrt(ab)`

Answer

To find the area of the greatest rectangle that can be inscribed in the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step 1: Understand the Geometry
The ellipse is symmetric about both the x-axis and y-axis. Therefore, we can focus on the first quadrant and then multiply the area of the rectangle by 4 to account for all four quadrants.

### Step 2: Parametrize the Ellipse
Using the parametric equations for the ellipse, we can express the coordinates of a point on the ellipse as:
\[ x = a \cos \theta, \quad y = b \sin \theta \]
where \( \theta \) is the parameter that varies.

### Step 3: Determine the Rectangle's Dimensions
The rectangle inscribed in the ellipse will have its corners at:
\[ (a \cos \theta, b \sin \theta), \quad (a \cos \theta, -b \sin \theta), \quad (-a \cos \theta, b \sin \theta), \quad (-a \cos \theta, -b \sin \theta) \]
The width of the rectangle is \( 2a \cos \theta \) and the height is \( 2b \sin \theta \).

### Step 4: Calculate the Area of the Rectangle
The area \( A \) of the rectangle can be expressed as:
\[ A = \text{width} \times \text{height} = (2a \cos \theta)(2b \sin \theta) = 4ab \cos \theta \sin \theta \]
Using the double angle identity, we can rewrite this as:
\[ A = 2ab \sin(2\theta) \]

### Step 5: Maximize the Area
To find the maximum area, we need to maximize \( \sin(2\theta) \). The maximum value of \( \sin(2\theta) \) is 1, which occurs when \( 2\theta = \frac{\pi}{2} \) or \( \theta = \frac{\pi}{4} \).

### Step 6: Calculate the Maximum Area
Substituting \( \sin(2\theta) = 1 \) into the area formula gives:
\[ A_{\text{max}} = 2ab \cdot 1 = 2ab \]

### Final Result
Thus, the area of the greatest rectangle that can be inscribed in the ellipse is:
\[ \boxed{2ab} \]