Area of Triangle Formed by Ellipse Point and Focus

To find the area of $\Delta POS$, we first determine the equation of the given ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Given:

• A focus of the ellipse is $S(4, 0)$, which corresponds to $(ae, 0)$. Thus, $ae = 4$.
• The eccentricity $e = \frac{4}{5}$.

Using $ae = 4$: $a \left(\frac{4}{5}\right) = 4 \implies a = 5$

Using the relation $b^2 = a^2(1 - e^2)$: $b^2 = 5^2 \left(1 - \left(\frac{4}{5}\right)^2\right) = 25 \left(1 - \frac{16}{25}\right) = 25 \left(\frac{9}{25}\right) = 9$

Thus, the equation of the ellipse $E$ is: $\frac{x^2}{25} + \frac{y^2}{9} = 1$

Since the point $P(3, \alpha)$ lies on the ellipse $E$, its coordinates must satisfy the equation: $\frac{3^2}{25} + \frac{\alpha^2}{9} = 1$ $\frac{9}{25} + \frac{\alpha^2}{9} = 1$ $\frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25}$ $\alpha^2 = \frac{144}{25} \implies |\alpha| = \frac{12}{5}$

The vertices of $\Delta POS$ are $O(0, 0)$, $S(4, 0)$, and $P(3, \alpha)$.

The area of $\Delta POS$ can be calculated as: $\text{Area}(\Delta POS) = \frac{1}{2} \times \text{base} \times \text{height}$

Taking the segment $OS$ along the x-axis as the base, we have: $\text{Base } OS = 4$ $\text{Height} = |\alpha| = \frac{12}{5}$

Substituting these values into the area formula gives: $\text{Area}(\Delta POS) = \frac{1}{2} \times 4 \times \frac{12}{5} = \frac{24}{5}$

Thus, the area of $\Delta POS$ is equal to $\frac{24}{5}$.