The equations of the tangent and normal at po | vedclass

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(A) Given the equation of the ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$.The equation of the tangent at

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$\frac{x}{3} - \frac{y}{2} = 1, \frac{x}{2} + \frac{y}{3} = \frac{5}{6}$

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(A) Given the equation of the ellipse is $4x^2 + 9y^2 = 36$,which can be written as $\frac{x^2}{9} + \frac{y^2}{4} = 1$.The equation of the tangent at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.Substituting $(x_1, y_1) = (3, -2)$,$a^2 = 9$,and $b^2 = 4$:$\frac{3x}{9} + \frac{-2y}{4} = 1 \Rightarrow \frac{x}{3} - \frac{y}{2} = 1$.The slope of the tangent is $m_t = \frac{1/3}{1/2} = \frac{2}{3}$.The slope of the normal $m_n$ is $-\frac{1}{m_t} = -\frac{3}{2}$.The equation of the normal at $(3, -2)$ is $y - (-2) = -\frac{3}{2}(x - 3)$.$2(y + 2) = -3(x - 3)$ $\Rightarrow 2y + 4 = -3x + 9$ $\Rightarrow 3x + 2y = 5$.Dividing by $6$,we get $\frac{3x}{6} + \frac{2y}{6} = \frac{5}{6} \Rightarrow \frac{x}{2} + \frac{y}{3} = \frac{5}{6}$.

The equations of the tangent and normal at point $(3, -2)$ of the ellipse $4x^2 + 9y^2 = 36$ are:

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