If alpha x+beta y=109 is the equation of the | vedclass

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(C) The equation of a chord of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T = \frac{xx_1}{

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$58$

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(C) The equation of a chord of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T = \frac{xx_1}{a^2}+\frac{yy_1}{b^2}$ and $S_1 = \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$.Given the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and midpoint $\left(\frac{5}{2}, \frac{1}{2}\right)$,we have:$\frac{x(5/2)}{9}+\frac{y(1/2)}{4} = \frac{(5/2)^2}{9}+\frac{(1/2)^2}{4}-1$$\Rightarrow \frac{5x}{18}+\frac{y}{8} = \frac{25/4}{9}+\frac{1/4}{4}-1$$\Rightarrow \frac{5x}{18}+\frac{y}{8} = \frac{25}{36}+\frac{1}{16}-1$$\Rightarrow \frac{5x}{18}+\frac{y}{8} = \frac{100+9-144}{144} = \frac{-35}{144}$Wait,the standard formula is $T=S_1$ where $S_1 = \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$. Let us re-evaluate:$\frac{5x}{18}+\frac{y}{8} = \frac{25}{36}+\frac{1}{16} = \frac{100+9}{144} = \frac{109}{144}$Multiplying by $144$:$8(5x) + 18(y) = 109$$40x + 18y = 109$Comparing with $\alpha x + \beta y = 109$,we get $\alpha = 40$ and $\beta = 18$.Therefore,$\alpha + \beta = 40 + 18 = 58$.

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

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