Let P be a variable point on the ellipse x^2 | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{3} + \frac{y^2}{1} = 1$. Let the point $P$ be $(\sqrt{3} \cos 	heta, \sin 	heta)$. The perpendicular

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$6$

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(C) The equation of the ellipse is $\frac{x^2}{3} + \frac{y^2}{1} = 1$. Let the point $P$ be $(\sqrt{3} \cos 	heta, \sin 	heta)$. The perpendicular distance of $P$ from the line $x - y - 10 = 0$ is given by $d = \frac{|\sqrt{3} \cos 	heta - \sin 	heta - 10|}{\sqrt{1^2 + (-1)^2}} = \frac{|\sqrt{3} \cos 	heta - \sin 	heta - 10|}{\sqrt{2}}$. To maximize $d$, we need to maximize the expression $f(	heta) = \sqrt{3} \cos 	heta - \sin 	heta$. We can write $f(	heta) = 2(\frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta) = 2 \cos(	heta + 30^{\circ})$. The maximum value of $f(	heta)$ is $2$ and the minimum value is $-2$. For maximum distance, we consider the minimum value of $f(	heta)$, which is $-2$. Thus, $d_{max} = \frac{|-2 - 10|}{\sqrt{2}} = \frac{|-12|}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2}$.

Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$. Then the maximum perpendicular distance of $P$ from the line $x - y = 10$ is (in $\sqrt{2}$)

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