Consider the conic e x^2 + pi y^2 - 2 e^2 x - | vedclass

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(C) The given equation is $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.Rearranging the terms,we get $e(x^2 - 2ex) + \pi(y^2 - 2\pi y)

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$2 \sqrt{\pi}$

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(C) The given equation is $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.Rearranging the terms,we get $e(x^2 - 2ex) + \pi(y^2 - 2\pi y) = \pi e - e^3 - \pi^3$.Completing the square,$e(x^2 - 2ex + e^2) + \pi(y^2 - 2\pi y + \pi^2) = \pi e - e^3 - \pi^3 + e^3 + \pi^3$.This simplifies to $e(x - e)^2 + \pi(y - \pi)^2 = \pi e$.Dividing by $\pi e$,we get $\frac{(x - e)^2}{\pi} + \frac{(y - \pi)^2}{e} = 1$.This is the equation of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ where $a^2 = \pi$ and $b^2 = e$.Since $\pi > e$,the major axis is along the $x$-direction,so $a = \sqrt{\pi}$.For any point $P$ on an ellipse,the sum of the distances from the foci $P S_1 + P S_2$ is constant and equal to the length of the major axis,which is $2a$.Therefore,$P S_1 + P S_2 = 2 \sqrt{\pi}$.

Consider the conic $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.
Suppose $P$ is any point on the conic and $S_1, S_2$ are the foci of the conic,then the maximum value of $(P S_1 + P S_2)$ is

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Consider the conic $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.Suppose $P$ is any point on the conic and $S_1, S_2$ are the foci of the conic,then the maximum value of $(P S_1 + P S_2)$ is