If the chord of the ellipse frac{x^2}{4}+frac | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{9}=1$. The equation of the chord of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with m

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$\alpha+1=\beta$

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(B) The equation of the ellipse is $\frac{x^2}{4}+\frac{y^2}{9}=1$. The equation of the 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 $T = \frac{xx_1}{a^2}+\frac{yy_1}{b^2}-1$ and $S_1 = \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$. Substituting $(x_1, y_1) = (1, 1)$,$a^2=4$,and $b^2=9$: $\frac{x(1)}{4}+\frac{y(1)}{9}-1 = \frac{1^2}{4}+\frac{1^2}{9}-1$ $\frac{x}{4}+\frac{y}{9}-1 = \frac{1}{4}+\frac{1}{9}-1$ $\frac{x}{4}+\frac{y}{9} = \frac{1}{4}+\frac{1}{9} = \frac{9+4}{36} = \frac{13}{36}$ Multiplying by $4$: $x+\frac{4}{9}y = \frac{13}{9}$ Comparing this with $x+\alpha y=\beta$,we get $\alpha = \frac{4}{9}$ and $\beta = \frac{13}{9}$. Then $\alpha+1 = \frac{4}{9}+1 = \frac{13}{9} = \beta$. Thus,$\alpha+1=\beta$.

If the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ having $(1,1)$ as its midpoint is $x+\alpha y=\beta$,then

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