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(C) The equation of a chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ bisected at point $(x_1, y_1)$ is given by $T = S_1$.Here,$S_1 = \f

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$x + 2y = 4$

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(C) The equation of a chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ bisected at point $(x_1, y_1)$ is given by $T = S_1$.Here,$S_1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$ and $T = \frac{xx_1}{a^2} + \frac{yy_1}{b^2} - 1$.Given ellipse: $\frac{x^2}{36} + \frac{y^2}{9} = 1$,so $a^2 = 36$ and $b^2 = 9$.Point $(x_1, y_1) = (2, 1)$.$S_1 = \frac{2^2}{36} + \frac{1^2}{9} - 1 = \frac{4}{36} + \frac{1}{9} - 1 = \frac{1}{9} + \frac{1}{9} - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$.$T = \frac{x(2)}{36} + \frac{y(1)}{9} - 1 = \frac{x}{18} + \frac{y}{9} - 1$.Equating $T = S_1$:$\frac{x}{18} + \frac{y}{9} - 1 = -\frac{7}{9}$.Multiply by $18$:$x + 2y - 18 = -14$.$x + 2y = 4$.

What is the equation of the chord of the ellipse $\frac{x^2}{36} + \frac{y^2}{9} = 1$ that is bisected at the point $(2, 1)$?

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