If $\alpha x + \beta y = 109$ is the equation of the chord of the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$, whose mid point is $\left(\dfrac{5}{2}, \dfrac{1}{2}\right)$, then $\alpha + \beta$ is equal to
Step-by-Step Solution
Key Concept: Apply $T=S_1$ with midpoint $(\tfrac{5}{2},\tfrac{1}{2})$ on the ellipse $\tfrac{x^2}{9}+\tfrac{y^2}{4}=1$, then scale the result to match the form $\alpha x+\beta y=109$ and read off $\alpha$ and $\beta$.
$T=S_1$: $\dfrac{\tfrac{5}{2}x}{9}+\dfrac{\tfrac{1}{2}y}{4}=\dfrac{(\tfrac{5}{2})^2}{9}+\dfrac{(\tfrac{1}{2})^2}{4}$, i.e., $\dfrac{5x}{18}+\dfrac{y}{8}=\dfrac{25}{36}+\dfrac{1}{16}=\dfrac{100+9}{144}=\dfrac{109}{144}$. Multiplying by $144$: $40x+18y=109$. So $\alpha=40$, $\beta=18$, $\alpha+\beta=58$.
Correct Answer: 1