Two tangents are drawn from the point `((25)/(3),0)` to the ellipse `(x^(2))/(25)+(y^(2))/(16)=1` if area of the triangle formed by the tangents with the chord of contact is equal to A then find the value of A.

Answer

To solve the problem, we need to find the area of the triangle formed by the tangents drawn from the point \((\frac{25}{3}, 0)\) to the ellipse \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) and the chord of contact.

### Step-by-Step Solution:

1. **Identify the Ellipse and the Point:**
The given ellipse is \(\frac{x^2}{25} + \frac{y^2}{16} = 1\). The point from which the tangents are drawn is \((\frac{25}{3}, 0)\).

2. **Find the Equation of the Chord of Contact:**
The equation of the chord of contact from the point \((x_1, y_1)\) to the ellipse is given by:
\[ \frac{xx_1}{25} + \frac{yy_1}{16} = 1 \]
Substituting \(x_1 = \frac{25}{3}\) and \(y_1 = 0\):
\[ \frac{x \cdot \frac{25}{3}}{25} + \frac{y \cdot 0}{16} = 1 \]
Simplifying this gives:
\[ \frac{x}{3} = 1 \quad \Rightarrow \quad x = 3 \]
Thus, the chord of contact is the vertical line \(x = 3\).

3. **Find Points of Intersection with the Ellipse:**
To find the points where this line intersects the ellipse, substitute \(x = 3\) into the ellipse equation:
\[ \frac{3^2}{25} + \frac{y^2}{16} = 1 \]
This simplifies to:
\[ \frac{9}{25} + \frac{y^2}{16} = 1 \]
Rearranging gives:
\[ \frac{y^2}{16} = 1 - \frac{9}{25} = \frac{16}{25} \]
Thus:
\[ y^2 = 16 \cdot \frac{16}{25} = \frac{256}{25} \quad \Rightarrow \quad y = \pm \frac{16}{5} \]
Therefore, the points of intersection are \((3, \frac{16}{5})\) and \((3, -\frac{16}{5})\).

4. **Determine the Area of the Triangle:**
The area \(A\) of the triangle formed by the tangents and the chord of contact can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the distance between the points of intersection on the y-axis, which is:
\[ \text{base} = \frac{16}{5} - \left(-\frac{16}{5}\right) = \frac{16}{5} + \frac{16}{5} = \frac{32}{5} \]
The height of the triangle is the horizontal distance from the point \((\frac{25}{3}, 0)\) to the line \(x = 3\):
\[ \text{height} = \left|\frac{25}{3} - 3\right| = \left|\frac{25}{3} - \frac{9}{3}\right| = \left|\frac{16}{3}\right| = \frac{16}{3} \]
Thus, the area \(A\) is:
\[ A = \frac{1}{2} \times \frac{32}{5} \times \frac{16}{3} = \frac{32 \times 16}{30} = \frac{512}{30} = \frac{256}{15} \]

### Final Answer:
The value of \(A\) is \(\frac{256}{15}\).