The equation of the chord of the ellipse $4 x^{2} + 9 y^{2} = 36$ having $(3, 2)$ as midpoint is

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$3 x + y = 11$

$3 x + 2 y = 13$

$2 x + 3 y = 12$

$2 x + 3 y = 6$

answer is B.

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Detailed Solution

$4 x^{2} + 9 y^{2} = 36 \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \Rightarrow a^{2} = 9, b^{2} = 4$

Midpoint $(3, 2) = (x_{1}, y_{1})$

Equation of chord $S_{1} = S_{11}$

$\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = \frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}}$

$\frac{x .3}{9} + \frac{y .2}{4} = \frac{9}{9} + \frac{4}{4}, \frac{3 x}{9} + \frac{2 y}{4} = 2$

$\frac{x}{3} + \frac{y}{2} = 2 \Rightarrow 2 x + 3 y = 12$

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