At which point on the ellipse 4x^2 + 9y^2 = 1 | vedclass

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(B) The equation of the ellipse is $4x^2 + 9y^2 = 1$. Let the point of tangency be $(x_1, y_1)$. The equation of the tangent at $(x_1, y_1)$ is $4xx_1

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of the ellipse is $4x^2 + 9y^2 = 1$. 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 $m = -\frac{4x_1}{9y_1}$. The given line is $8x = 9y$,which can be written as $y = \frac{8}{9}x$. The slope of this line is $\frac{8}{9}$. Since the tangent is parallel to the line,their slopes are equal: $-\frac{4x_1}{9y_1} = \frac{8}{9}$ $\Rightarrow -\frac{x_1}{y_1} = 2$ $\Rightarrow x_1 = -2y_1$. Since $(x_1, y_1)$ lies on the ellipse,$4x_1^2 + 9y_1^2 = 1$. Substituting $x_1 = -2y_1$: $4(-2y_1)^2 + 9y_1^2 = 1 $ $\Rightarrow 4(4y_1^2) + 9y_1^2 = 1 $ $\Rightarrow 16y_1^2 + 9y_1^2 = 1 $ $\Rightarrow 25y_1^2 = 1 $ $\Rightarrow y_1^2 = \frac{1}{25} $ $\Rightarrow y_1 = \pm \frac{1}{5}$. 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 $\left( \frac{2}{5}, -\frac{1}{5} \right)$.

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

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