If the midpoint of a chord of the ellipse $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ is $$(\sqrt{2},4/3)$$, and the length of the chord is $$\frac{2sqrt{\alpha}}{3}$$, then $$\alpha$$ is :

We are given the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ with midpoint of a chord at $$(\sqrt{2}, \frac{4}{3})$$.

We start by finding the equation of the chord with midpoint $$(h, k)$$ in the standard form of an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, which is given by $$\frac{xh}{a^2} + \frac{yk}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$$.

Substituting $$a^2 = 9$$, $$b^2 = 4$$, $$h = \sqrt{2}$$, and $$k = \frac{4}{3}$$ into this formula yields $$\frac{\sqrt{2}x}{9} + \frac{4y}{3 \cdot 4} = \frac{2}{9} + \frac{16}{36}$$, which simplifies to $$\frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{3}$$. Multiplying through by 9 gives $$\sqrt{2}x + 3y = 6$$, so $$y = \frac{6 - \sqrt{2}x}{3}$$.

Next, we find the endpoints of the chord by substituting this expression for $$y$$ into the ellipse equation: $$\frac{x^2}{9} + \frac{1}{4}\left(\frac{6 - \sqrt{2}x}{3}\right)^2 = 1$$, which is equivalent to $$\frac{x^2}{9} + \frac{(6 - \sqrt{2}x)^2}{36} = 1$$. Multiplying both sides by 36 leads to $$4x^2 + (6 - \sqrt{2}x)^2 = 36$$, and expanding gives $$4x^2 + 36 - 12\sqrt{2}x + 2x^2 = 36$$. This simplifies to $$6x^2 - 12\sqrt{2}x = 0$$ or $$6x(x - 2\sqrt{2}) = 0$$, so the solutions are $$x_1 = 0$$ and $$x_2 = 2\sqrt{2}$$.

When $$x = 0$$, we have $$y = \frac{6}{3} = 2$$, giving the point $$(0, 2)$$, and when $$x = 2\sqrt{2}$$, we get $$y = \frac{6 - \sqrt{2} \cdot 2\sqrt{2}}{3} = \frac{6 - 4}{3} = \frac{2}{3}$$, giving the point $$(2\sqrt{2}, \frac{2}{3})$$.

To verify the midpoint, we compute $$\left(\frac{0 + 2\sqrt{2}}{2}, \frac{2 + \frac{2}{3}}{2}\right) = \left(\sqrt{2}, \frac{4}{3}\right)$$, which matches the given midpoint.

Then the length of the chord is $$\sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2} = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{72 + 16}{9}} = \frac{\sqrt{88}}{3} = \frac{2\sqrt{22}}{3}$$.

Comparing this result with the given expression $$\frac{2\sqrt{\alpha}}{3}$$ shows that $$\alpha = 22$$, so the correct answer is 22.