The area of the parallelogram inscribed in the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1`, whose diaonals are the conjugate diameters of the ellipse is given by

Options

A. (a)2ab
B. (b)3ab
C. (c)4ab
D. (d)5ab

Answer

To find the area of the parallelogram inscribed in the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), whose diagonals are the conjugate diameters of the ellipse, we can follow these steps:

### Step 1: Understand the Conjugate Diameters
The conjugate diameters of the ellipse are lines that are perpendicular to each other and pass through the center of the ellipse. For the ellipse given, the equations of the conjugate diameters can be expressed as:
\[ y = \frac{b}{a} x \quad \text{and} \quad y = -\frac{b}{a} x \]

### Step 2: Find the Points of Intersection
To find the points where these lines intersect the ellipse, we substitute \( y = \frac{b}{a} x \) into the ellipse equation:
\[ \frac{x^2}{a^2} + \frac{\left(\frac{b}{a} x\right)^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{a^2} + \frac{b^2 x^2}{a^2 b^2} = 1 \]
\[ \frac{x^2}{a^2} + \frac{x^2}{a^2} = 1 \]
\[ \frac{2x^2}{a^2} = 1 \implies x^2 = \frac{a^2}{2} \implies x = \pm \frac{a}{\sqrt{2}} \]

### Step 3: Calculate the Corresponding y-coordinates
Using \( x = \frac{a}{\sqrt{2}} \) to find \( y \):
\[ y = \frac{b}{a} \cdot \frac{a}{\sqrt{2}} = \frac{b}{\sqrt{2}} \]
Thus, the points of intersection are:
\[ P\left(-\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right), \quad Q\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right) \]

### Step 4: Find the Other Two Points
For the other conjugate diameter, substituting \( y = -\frac{b}{a} x \) into the ellipse equation gives:
\[ \frac{x^2}{a^2} + \frac{\left(-\frac{b}{a} x\right)^2}{b^2} = 1 \]
This leads to the same \( x \) values:
\[ R\left(-\frac{a}{\sqrt{2}}, -\frac{b}{\sqrt{2}}\right), \quad S\left(\frac{a}{\sqrt{2}}, -\frac{b}{\sqrt{2}}\right) \]

### Step 5: Calculate the Area of the Parallelogram
The area \( A \) of the parallelogram formed by the points \( P, Q, R, S \) can be calculated using the formula:
\[ A = \text{Base} \times \text{Height} \]
The base can be taken as the distance between points \( P \) and \( Q \):
\[ \text{Base} = PQ = \left| \frac{a}{\sqrt{2}} - \left(-\frac{a}{\sqrt{2}}\right) \right| = \frac{2a}{\sqrt{2}} = \sqrt{2}a \]
The height can be taken as the distance between points \( R \) and \( S \):
\[ \text{Height} = RS = \left| \frac{b}{\sqrt{2}} - \left(-\frac{b}{\sqrt{2}}\right) \right| = \frac{2b}{\sqrt{2}} = \sqrt{2}b \]
Thus, the area \( A \) is:
\[ A = \text{Base} \times \text{Height} = \sqrt{2}a \times \sqrt{2}b = 2ab \]

### Final Answer
The area of the parallelogram inscribed in the ellipse is:
\[ \boxed{2ab} \]