On the ellipse frac{x^2}{18} + frac{y^2}{8} = | vedclass

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(A) Let the point $M$ on the ellipse be $(a \cos 	heta, b \sin 	heta)$,where $a = \sqrt{18} = 3\sqrt{2}$ and $b = \sqrt{8} = 2\sqrt{2}$.So,$M = (3\s

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$(-3, 2)$

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(A) Let the point $M$ on the ellipse be $(a \cos 	heta, b \sin 	heta)$,where $a = \sqrt{18} = 3\sqrt{2}$ and $b = \sqrt{8} = 2\sqrt{2}$.So,$M = (3\sqrt{2} \cos 	heta, 2\sqrt{2} \sin 	heta)$.The shortest distance from a point on the ellipse to a line occurs where the tangent at that point is parallel to the given line.The slope of the given line $2x - 3y + 25 = 0$ is $m = \frac{2}{3}$.The slope of the tangent at $M$ is given by $-\frac{b^2 x_0}{a^2 y_0} = -\frac{8(3\sqrt{2} \cos 	heta)}{18(2\sqrt{2} \sin 	heta)} = -\frac{24 \cos 	heta}{36 \sin 	heta} = -\frac{2}{3} \cot 	heta$.Equating the slopes: $-\frac{2}{3} \cot 	heta = \frac{2}{3}$ $\Rightarrow \cot 	heta = -1$ $\Rightarrow 	an 	heta = -1$.For the point to be nearest to the line,we choose $	heta = \frac{3\pi}{4}$ (in the second quadrant where $x$ is negative and $y$ is positive).$x = 3\sqrt{2} \cos(\frac{3\pi}{4}) = 3\sqrt{2} 	imes (-\frac{1}{\sqrt{2}}) = -3$.$y = 2\sqrt{2} \sin(\frac{3\pi}{4}) = 2\sqrt{2} 	imes (\frac{1}{\sqrt{2}}) = 2$.Thus,the point $M$ is $(-3, 2)$.

On the ellipse $\frac{x^2}{18} + \frac{y^2}{8} = 1$,the point $M$ nearest to the line $2x - 3y + 25 = 0$ is

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