If the midpoint of a chord of the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ is $\left(\sqrt{2}, \dfrac{4}{3}\right)$, and the length of the chord is $\dfrac{2\sqrt{\alpha}}{3}$, then $\alpha$ is
Step-by-Step Solution
Key Concept: Use $T=S_1$ to get the chord equation, substitute back into the ellipse equation to find the two endpoints, then compute the chord length and match to $\tfrac{2\sqrt{\alpha}}{3}$.
$T=S_1$: $\dfrac{\sqrt{2}\,x}{9}+\dfrac{y}{3}=\dfrac{2}{9}+\dfrac{4}{9}$, giving $\sqrt{2}\,x+3y=6$, i.e., $x=\tfrac{6-3y}{\sqrt{2}}$. Substituting into the ellipse gives $3y^2-8y+4=0$, so $y=2$ or $y=\tfrac{2}{3}$. Endpoints: $(0,2)$ and $(2\sqrt{2},\tfrac{2}{3})$. Chord length $=\sqrt{(2\sqrt{2})^2+(\tfrac{2}{3}-2)^2}=\sqrt{8+\tfrac{16}{9}}=\sqrt{\tfrac{88}{9}}=\tfrac{2\sqrt{22}}{3}$. Hence $\alpha=22$.
Correct Answer: 2