On the ellipse 4x^2 + 9y^2 = 1,the points at | vedclass

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(D) The given ellipse is $4x^2 + 9y^2 = 1$,which can be written as $\frac{x^2}{1/4} + \frac{y^2}{1/9} = 1$.The slope of the line $8x = 9y$ is $m = \fr

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$B$ and $C$ both

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(D) The given ellipse is $4x^2 + 9y^2 = 1$,which can be written as $\frac{x^2}{1/4} + \frac{y^2}{1/9} = 1$.The slope of the line $8x = 9y$ is $m = \frac{8}{9}$.Let the point of tangency be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $4xx_1 + 9yy_1 = 1$.The slope of this tangent is $-\frac{4x_1}{9y_1}$.Since the tangent is parallel to the line,we have $-\frac{4x_1}{9y_1} = \frac{8}{9}$,which simplifies to $x_1 = -2y_1$.Substituting $x_1 = -2y_1$ into the ellipse equation $4x_1^2 + 9y_1^2 = 1$:$4(-2y_1)^2 + 9y_1^2 = 1$$16y_1^2 + 9y_1^2 = 1$$25y_1^2 = 1 \implies y_1 = \pm \frac{1}{5}$.If $y_1 = \frac{1}{5}$,then $x_1 = -2(\frac{1}{5}) = -\frac{2}{5}$.If $y_1 = -\frac{1}{5}$,then $x_1 = -2(-\frac{1}{5}) = \frac{2}{5}$.Thus,the points are $\left( -\frac{2}{5}, \frac{1}{5} \right)$ and $\left( \frac{2}{5}, -\frac{1}{5} \right)$.

On the ellipse $4x^2 + 9y^2 = 1$,the points at which the tangents are parallel to the line $8x = 9y$ are

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