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

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(D) Given the equation: $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$. Let $P = (x, y)$,$A = (2, 0)$,and $B = (-2, 0)$. The equation represents the sum of

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Line segment

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(D) Given the equation: $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$. Let $P = (x, y)$,$A = (2, 0)$,and $B = (-2, 0)$. The equation represents the sum of distances from point $P$ to points $A$ and $B$,which is $PA + PB = 4$. The distance between $A(2, 0)$ and $B(-2, 0)$ is $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$. Since $PA + PB = AB$,the point $P$ must lie on the line segment connecting $A$ and $B$. Given the condition $-2 Thus,the equation represents a line segment. Hence,option $D$ is correct.

The equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$,where $-2 < x < 2$,represents a

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