If x+ y=109 is the equation of the chord of the ellipse x^29+y^24=1, whose midpoint is (52…

Question

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 midpoint is $\left(\dfrac52,\dfrac12\right)$, then $\alpha+\beta$ is equal to:

Accepted Answer

$T=S_1$: $\dfrac{(5/2)x}{9}+\dfrac{(1/2)y}{4}=\dfrac{25/4}{9}+\dfrac{1/4}{4} \Rightarrow \dfrac{5x}{18}+\dfrac{y}{8}=\dfrac{109}{144} \Rightarrow 40x+18y=109.$

So $\alpha=40,\ \beta=18,\ \alpha+\beta=58$.

Solutions · 1