The equation sqrt{(x - 2)^2 + y^2} + sqrt{(x | vedclass

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(D) Let $P(x, y)$ be a point in the plane. The given equation is $PF_1 + PF_2 = 4$,where $F_1 = (2, 0)$ and $F_2 = (-2, 0)$.Here,the distance between

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$A$ line segment on the $x$-axis

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(D) Let $P(x, y)$ be a point in the plane. The given equation is $PF_1 + PF_2 = 4$,where $F_1 = (2, 0)$ and $F_2 = (-2, 0)$.Here,the distance between the foci $F_1$ and $F_2$ is $d(F_1, F_2) = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$.Since the sum of the distances from $P$ to two fixed points $F_1$ and $F_2$ is equal to the distance between the points themselves $(PF_1 + PF_2 = F_1F_2)$,the point $P$ must lie on the line segment connecting $F_1$ and $F_2$.Thus,the equation represents the line segment on the $x$-axis between $x = -2$ and $x = 2$,which is a degenerate case of an ellipse where the minor axis is $0$.

The equation $\sqrt{(x - 2)^2 + y^2} + \sqrt{(x + 2)^2 + y^2} = 4$ represents:

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