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 $\left(\frac{5}{2}, \frac{1}{2}\right)$, then $\alpha+\beta$ is equal to :

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Aicharya · Accepted Answer

Equation of chord T $=\mathrm{S}_1$
$\begin{aligned}
& \frac{5}{2}\left(\frac{x}{9}\right)+\frac{1}{2}\left(\frac{\mathrm{y}}{4}\right)=\frac{25}{36}+\frac{1}{16} \\ & \Rightarrow \frac{5 \mathrm{x}}{18}+\frac{\mathrm{y}}{8}=\frac{100+9}{144}=\frac{109}{144} \\ & \Rightarrow 40 \mathrm{x}+18 \mathrm{y}=109 \\ & \Rightarrow \alpha=40, \beta=18 \\ & \Rightarrow \alpha+\beta=58
\end{aligned}$

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…

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