The equation of the normal at the point (0, 3 | vedclass

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(D) Given the ellipse equation: $9x^2 + 5y^2 = 45$.Dividing by $45$,we get $\frac{x^2}{5} + \frac{y^2}{9} = 1$.Here,$a^2 = 5$ and $b^2 = 9$.The point

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$y$-axis

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(D) Given the ellipse equation: $9x^2 + 5y^2 = 45$.Dividing by $45$,we get $\frac{x^2}{5} + \frac{y^2}{9} = 1$.Here,$a^2 = 5$ and $b^2 = 9$.The point $(x_1, y_1) = (0, 3)$ lies on the ellipse.The equation of the normal at $(x_1, y_1)$ is given by $\frac{a^2(x - x_1)}{x_1} = \frac{b^2(y - y_1)}{y_1}$.Substituting the values: $\frac{5(x - 0)}{0} = \frac{9(y - 3)}{3}$.This implies $\frac{5x}{0} = 3(y - 3)$.For the expression to be defined,$x$ must be $0$.Thus,the equation of the normal is $x = 0$,which represents the $y$-axis.

The equation of the normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

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