If $$\alpha x+ \beta y = 109$$ is the equation of the chord of the ellipse $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$, whose mid point is $$(\frac{5}{2},\frac{1}{2})$$ , then $$\alpha + \beta$$
is equal to :

For the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$, the equation of the chord with midpoint $$(h, k)$$ is given by:

$$ \frac{xh}{9} + \frac{yk}{4} = \frac{h^2}{9} + \frac{k^2}{4} $$

Given midpoint $$\left(\frac{5}{2}, \frac{1}{2}\right)$$:

$$ \frac{x \cdot \frac{5}{2}}{9} + \frac{y \cdot \frac{1}{2}}{4} = \frac{\left(\frac{5}{2}\right)^2}{9} + \frac{\left(\frac{1}{2}\right)^2}{4} $$

$$ \frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16} $$

RHS: $$\frac{25}{36} + \frac{1}{16} = \frac{400 + 36}{576} = \frac{436}{576} = \frac{109}{144}$$

LHS: $$\frac{5x}{18} + \frac{y}{8}$$

Multiplying through by 144:

$$ 40x + 18y = 109 $$

Comparing with $$\alpha x + \beta y = 109$$:

$$\alpha = 40, \quad \beta = 18$$

$$\alpha + \beta = 40 + 18 = 58$$

The correct answer is Option 1: 58.