On the ellipse 4x^{2} + 9y^{2} = 1, the point | vedclass

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(B, D) The given equation of the ellipse is $4x^{2} + 9y^{2} = 1$.Differentiating with respect to $x$, we get $8x + 18yy' = 0$, which implies $y' = -\

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$\left(\frac{2}{5}, -\frac{1}{5}\right)$

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(B, D) The given equation of the ellipse is $4x^{2} + 9y^{2} = 1$.Differentiating with respect to $x$, we get $8x + 18yy' = 0$, which implies $y' = -\frac{8x}{18y} = -\frac{4x}{9y}$.The slope of the line $8x = 9y$ is $y = \frac{8}{9}x$, so the slope $m = \frac{8}{9}$.Since the tangents are parallel to the line, their slopes must be equal: $-\frac{4x}{9y} = \frac{8}{9}$.This simplifies to $-4x = 8y$, or $x = -2y$.Substituting $x = -2y$ into the ellipse equation: $4(-2y)^{2} + 9y^{2} = 1$.$4(4y^{2}) + 9y^{2} = 1$ $\Rightarrow 16y^{2} + 9y^{2} = 1$ $\Rightarrow 25y^{2} = 1$.Thus, $y^{2} = \frac{1}{25}$, which gives $y = \pm \frac{1}{5}$.If $y = \frac{1}{5}$, then $x = -2(\frac{1}{5}) = -\frac{2}{5}$.If $y = -\frac{1}{5}$, then $x = -2(-\frac{1}{5}) = \frac{2}{5}$.The required 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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