Let a focus of the ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ be $$S(4, 0)$$ and its eccentricity be $$\frac{4}{5}$$. If $$P(3, \alpha)$$ lies on  $$E$$ and $$O$$ is the origin, then the area of $$\triangle POS$$ is equal to:

To find the area of $$\triangle POS$$, we need to determine the coordinates of point $$P(3, \alpha)$$ by first finding the constants $$a^2$$ and $$b^2$$ for the ellipse.

The standard equation of the ellipse is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$.

We are given:

• Focus ($$S$$): $$(4, 0)$$. In an ellipse, the focus is at $$(ae, 0)$$. Thus, $$ae = 4$$.
• Eccentricity ($$e$$): $$\frac{4}{5}$$.
• $$O(0, 0)$$
• $$S(4, 0)$$
• $$P(3, \frac{12}{5})$$
• Base ($$OS$$): $$4$$ units (distance from $$0$$ to $$4$$ on the $$x$$-axis).
• Height ($$h$$): The $$y$$-coordinate of point $$P$$, which is $$\frac{12}{5}$$.

Using $$ae = 4$$:

$$a \left( \frac{4}{5} \right) = 4 \implies a = 5 \implies a^2 = 25$$

Now, use the relation $$b^2 = a^2(1 - e^2)$$:

$$b^2 = 25 \left( 1 - \left(\frac{4}{5}\right)^2 \right)$$

$$b^2 = 25 \left( 1 - \frac{16}{25} \right) = 25 \left( \frac{9}{25} \right) = 9$$

The equation of the ellipse is:

$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Point $$P(3, \alpha)$$ lies on the ellipse. Substitute $$x = 3$$ and $$y = \alpha$$ into the equation:

$$\frac{3^2}{25} + \frac{\alpha^2}{9} = 1$$

$$\frac{9}{25} + \frac{\alpha^2}{9} = 1$$

$$\frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25}$$

$$\alpha^2 = \frac{16 \cdot 9}{25} \implies \alpha = \pm \frac{4 \cdot 3}{5} = \pm \frac{12}{5}$$

Since we are looking for the area of a triangle, we can take the absolute value: $$|\alpha| = \frac{12}{5}$$.

We have the coordinates of the three vertices:

Since the base $$OS$$ lies on the $$x$$-axis, the calculation is straightforward:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

$$\text{Area} = \frac{1}{2} \times 4 \times \frac{12}{5}$$

$$\text{Area} = 2 \times \frac{12}{5} = \frac{24}{5}$$

Conclusion:

The area of $$\triangle POS$$ is 24/5.

Correct Option: C