At which points on the ellipse 4x^2 + 9y^2 = | vedclass

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(B) Given the equation of the ellipse: $4x^2 + 9y^2 = 1$  Differentiating with respect to $x$: $8x + 18y \frac{dy}{dx} = 0$  $\frac{dy}{dx} = -\frac{8

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$\left( -\frac{2}{5}, \frac{1}{5} ight) 	ext{ or } \left( \frac{2}{5}, -\frac{1}{5} ight)$

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(B) Given the equation of the ellipse: $4x^2 + 9y^2 = 1$  Differentiating with respect to $x$: $8x + 18y \frac{dy}{dx} = 0$  $\frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}$  The given line is $8x = 9y$,which can be written as $y = \frac{8}{9}x$.  The slope of this line is $m = \frac{8}{9}$.  Since the tangent is parallel to the line,their slopes must be equal:  $-\frac{4x}{9y} = \frac{8}{9}$  $-4x = 8y \implies x = -2y$  Substitute $x = -2y$ into the ellipse equation:  $4(-2y)^2 + 9y^2 = 1$  $4(4y^2) + 9y^2 = 1$  $16y^2 + 9y^2 = 1$  $25y^2 = 1 \implies y^2 = \frac{1}{25} \implies y = \pm \frac{1}{5}$  If $y = \frac{1}{5}$,then $x = -2(\frac{1}{5}) = -\frac{2}{5}$.  If $y = -\frac{1}{5}$,then $x = -2(-\frac{1}{5}) = \frac{2}{5}$.  Thus,the points are $\left( -\frac{2}{5}, \frac{1}{5} \right)$ and $\left( \frac{2}{5}, -\frac{1}{5} \right)$.

At which points on the ellipse $4x^2 + 9y^2 = 1$ is the tangent line parallel to the line $8x = 9y$?

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