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(B) The equation of the ellipse is $\frac{x^2}{100} + \frac{y^2}{25} = 1$. The equation of the tangent at point $(x_1, y_1) = (-8, 3)$ is given by $\f

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$\left(0, \frac{25}{3}\right)$

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(B) The equation of the ellipse is $\frac{x^2}{100} + \frac{y^2}{25} = 1$. The equation of the tangent at point $(x_1, y_1) = (-8, 3)$ is given by $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$. Substituting the values: $\frac{x(-8)}{100} + \frac{y(3)}{25} = 1$. Simplifying the equation: $-\frac{2x}{25} + \frac{3y}{25} = 1$,which gives $-2x + 3y = 25$. To find the point where the particle crosses the $Y$-axis,we set $x = 0$. Substituting $x = 0$ into the tangent equation: $3y = 25$,which gives $y = \frac{25}{3}$. Thus,the point is $\left(0, \frac{25}{3}\right)$.

$A$ particle is travelling in a clockwise direction on the ellipse $\frac{x^2}{100} + \frac{y^2}{25} = 1$. If the particle leaves the ellipse at the point $(-8, 3)$ and travels along the tangent to the ellipse at that point,then the point where the particle crosses the $Y$-axis is:

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