Find the equation of the tangent to the ellip | vedclass

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(A) The given equation of the ellipse is $4x^2 + 9y^2 = 36$.Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Comparing this with the standa

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$\frac{x}{3} - \frac{y}{2} = 1$

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(A) The given equation of the ellipse is $4x^2 + 9y^2 = 36$.Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$.The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$ is given by $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.Substituting $(x_1, y_1) = (3, -2)$,$a^2 = 9$,and $b^2 = 4$ into the formula:$\frac{x(3)}{9} + \frac{y(-2)}{4} = 1$$\frac{3x}{9} - \frac{2y}{4} = 1$$\frac{x}{3} - \frac{y}{2} = 1$.

Find the equation of the tangent to the ellipse $4x^2 + 9y^2 = 36$ at the point $(3, -2)$.

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