The area of the parallelogram formed by the tangents at the points whose eccentric angles are `theta, theta +(pi)/(2), theta +pi, theta +(3pi)/(2)` on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =1` is

Options

A. ab
B. 4ab
C. 3ab
D. 2ab

Explanation

To find the area of the parallelogram formed by the tangents at the given points on the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can follow these steps:

### Step 1: Identify the points on the ellipse
The eccentric angles given are:
- \(\theta\)
- \(\theta + \frac{\pi}{2}\)
- \(\theta + \pi\)
- \(\theta + \frac{3\pi}{2}\)

Using the parametric equations of the ellipse:
- \(x = a \cos(\theta)\)
- \(y = b \sin(\theta)\)

The points corresponding to the given eccentric angles are:
1. For \(\theta\): \((a \cos(\theta), b \sin(\theta))\)
2. For \(\theta + \frac{\pi}{2}\): \((a \cos(\theta + \frac{\pi}{2}), b \sin(\theta + \frac{\pi}{2})) = (-b \sin(\theta), a \cos(\theta))\)
3. For \(\theta + \pi\): \((a \cos(\theta + \pi), b \sin(\theta + \pi)) = (-a \cos(\theta), -b \sin(\theta))\)
4. For \(\theta + \frac{3\pi}{2}\): \((a \cos(\theta + \frac{3\pi}{2}), b \sin(\theta + \frac{3\pi}{2})) = (b \sin(\theta), -a \cos(\theta))\)

### Step 2: Find the equations of the tangents
The equation of the tangent to the ellipse at point \((x_0, y_0)\) is given by:
\[ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1 \]

Using the points identified:
1. Tangent at \((a \cos(\theta), b \sin(\theta))\):
\[ \frac{xx_0}{a^2} + \frac{yy_0}{b^2} = \frac{x(a \cos(\theta))}{a^2} + \frac{y(b \sin(\theta))}{b^2} = 1 \]

2. Tangent at \((-b \sin(\theta), a \cos(\theta))\):
\[ \frac{x(-b \sin(\theta))}{a^2} + \frac{y(a \cos(\theta))}{b^2} = 1 \]

3. Tangent at \((-a \cos(\theta), -b \sin(\theta))\):
\[ \frac{x(-a \cos(\theta))}{a^2} + \frac{y(-b \sin(\theta))}{b^2} = 1 \]

4. Tangent at \((b \sin(\theta), -a \cos(\theta))\):
\[ \frac{x(b \sin(\theta))}{a^2} + \frac{y(-a \cos(\theta))}{b^2} = 1 \]

### Step 3: Determine the area of the parallelogram
The area of the parallelogram formed by these tangents can be determined by recognizing that the points correspond to the vertices of a rectangle. The lengths of the sides of the rectangle are:
- Length along the major axis: \(2a\)
- Length along the minor axis: \(2b\)

Thus, the area \(A\) of the parallelogram (which is a rectangle in this case) is given by:
\[ A = \text{Length} \times \text{Width} = 2a \times 2b = 4ab \]

### Final Answer
The area of the parallelogram formed by the tangents at the specified points is:
\[ \boxed{4ab} \]