The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $$\frac{x^2}{9} + \frac{y^2}{5} = 1$$, is

The given equation of the ellipse is $$\frac{x^2}{9} + \frac{y^2}{5} = 1$$

Comparing this with the standard form $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$: $$a^2 = 9 \implies a = 3$$, $$b^2 = 5$$

The eccentricity ($$e$$) is given by $$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$

The four endpoints of the latus rectum for an ellipse are $$(\pm ae, \pm \frac{b^2}{a})$$.

The point in the first quadrant is $$P(2, \frac{5}{3})$$

The equation of a tangent to the ellipse at $$(x_1, y_1)$$ is $$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$.

Substituting $$P(2, \frac{5}{3})$$: $$\frac{x(2)}{9} + \frac{y(5/3)}{5} = 1 \implies \frac{2x}{9} + \frac{y}{3} = 1$$

$$2x + 3y = 9$$

By symmetry, the four tangents forming the quadrilateral are $$2x \pm 3y = \pm 9$$. This figure is a rhombus.

$$\text{Area} = 4 \times \left( \frac{1}{2} \times x_{intercept} \times y_{intercept} \right) = 2 \times \frac{9}{2} \times 3$$

$$\text{Area} = 27 \text{ sq. units}$$