1. Use the midpoint form of a chord of an ellipse

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ the chord whose midpoint is $(x_1,y_1)$ has equation $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}.$

Here, $\frac{x^2}{9}+\frac{y^2}{4}=1,$ so $a^2=9,\qquad b^2=4,$ and the midpoint is $\left(\frac52,\frac12\right).$

So the chord equation is $\frac{x\cdot \frac52}{9}+\frac{y\cdot \frac12}{4}=\frac{\left(\frac52\right)^2}{9}+\frac{\left(\frac12\right)^2}{4}.$

2. Simplify the equation

Left side: $\frac{5x}{18}+\frac{y}{8}.$

Right side: $\frac{25/4}{9}+\frac{1/4}{4}=\frac{25}{36}+\frac{1}{16}.$ Taking LCM $144$, $\frac{25}{36}=\frac{100}{144},\qquad \frac{1}{16}=\frac{9}{144},$ so $\frac{25}{36}+\frac{1}{16}=\frac{109}{144}.$

Thus the chord is $\frac{5x}{18}+\frac{y}{8}=\frac{109}{144}.$

3. Convert to the given form

Multiply by $144$: $144\cdot \frac{5x}{18}+144\cdot \frac{y}{8}=109.$ So, $40x+18y=109.$

Hence, $\alpha=40,\qquad \beta=18.$ Therefore, $\alpha+\beta=40+18=58.$

4. Check the options

The correct option is $\boxed{58}$ which is Option D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D (58)

So the derived answer agrees with the stored answer.