Tangent at a point on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
is drawn which cuts the coordinates axes at A and B the minimum area of the triangle OAB is ( O being origin )

Options

A. ab
B. `(a^(3)+b^(3)+ab)/(3)`
C. `a^(2)+b^(2)`
D. `(a^(2)+b^(2))/(4)`

Explanation

To solve the problem, we need to find the minimum area of triangle OAB formed by the tangent to the ellipse at a point, where O is the origin, A is the intersection with the x-axis, and B is the intersection with the y-axis.

### Step-by-Step Solution:

1. **Equation of the Ellipse**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

2. **Parametric Representation**:
For a point on the ellipse, we can use the parametric equations:
\[ x = a \cos \theta, \quad y = b \sin \theta \]
where \(\theta\) is the eccentric angle.

3. **Equation of the Tangent**:
The equation of the tangent to the ellipse at the point \((a \cos \theta, b \sin \theta)\) is given by:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]

4. **Finding Intersection with the X-axis (Point A)**:
To find the x-intercept (point A), set \(y = 0\):
\[ \frac{x \cos \theta}{a} = 1 \implies x = \frac{a}{\cos \theta} \]
Thus, the coordinates of point A are \(\left(\frac{a}{\cos \theta}, 0\right)\).

5. **Finding Intersection with the Y-axis (Point B)**:
To find the y-intercept (point B), set \(x = 0\):
\[ \frac{y \sin \theta}{b} = 1 \implies y = \frac{b}{\sin \theta} \]
Thus, the coordinates of point B are \(\left(0, \frac{b}{\sin \theta}\right)\).

6. **Area of Triangle OAB**:
The area \(A\) of triangle OAB can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is OA and the height is OB:
\[ \text{Area} = \frac{1}{2} \times \frac{a}{\cos \theta} \times \frac{b}{\sin \theta} \]
Simplifying this gives:
\[ \text{Area} = \frac{ab}{2} \cdot \frac{1}{\sin \theta \cos \theta} \]
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we can rewrite the area as:
\[ \text{Area} = \frac{ab}{\sin 2\theta} \]

7. **Finding Minimum Area**:
To minimize the area, we need to maximize \(\sin 2\theta\). The maximum value of \(\sin 2\theta\) is 1. Thus, the minimum area occurs when:
\[ \text{Minimum Area} = \frac{ab}{1} = ab \]

### Final Answer:
The minimum area of triangle OAB is:
\[ \text{Minimum Area} = ab \]