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(B) Let the point be $P(x, y)$. The given condition is $PA + PB = 4$.$\sqrt{(x-1)^2 + (y+1)^2} + \sqrt{(x+1)^2 + (y-1)^2} = 4$Squaring both sides:$(x-

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$3x^2 + 2xy + 3y^2 = 8$

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(B) Let the point be $P(x, y)$. The given condition is $PA + PB = 4$.$\sqrt{(x-1)^2 + (y+1)^2} + \sqrt{(x+1)^2 + (y-1)^2} = 4$Squaring both sides:$(x-1)^2 + (y+1)^2 + (x+1)^2 + (y-1)^2 + 2\sqrt{((x-1)^2 + (y+1)^2)((x+1)^2 + (y-1)^2)} = 16$$2(x^2 + y^2 + 2) + 2\sqrt{(x^2 + y^2 + 2 - 2x + 2y)(x^2 + y^2 + 2 + 2x - 2y)} = 16$$(x^2 + y^2 + 2) + \sqrt{(x^2 + y^2 + 2)^2 - (2x - 2y)^2} = 8$$\sqrt{(x^2 + y^2 + 2)^2 - 4(x - y)^2} = 8 - (x^2 + y^2 + 2)$Squaring again:$(x^2 + y^2 + 2)^2 - 4(x - y)^2 = (6 - (x^2 + y^2))^2$$(x^2 + y^2 + 2)^2 - (x^2 + y^2 - 6)^2 = 4(x - y)^2$Using $a^2 - b^2 = (a-b)(a+b)$:$((x^2 + y^2 + 2) - (x^2 + y^2 - 6))((x^2 + y^2 + 2) + (x^2 + y^2 - 6)) = 4(x^2 + y^2 - 2xy)$$(8)(2x^2 + 2y^2 - 4) = 4(x^2 + y^2 - 2xy)$$16(x^2 + y^2 - 2) = 4(x^2 + y^2 - 2xy)$$4x^2 + 4y^2 - 8 = x^2 + y^2 - 2xy$$3x^2 + 2xy + 3y^2 = 8$

If a point $P$ moves such that the sum of the distances from $P$ to the points $A(1, -1)$ and $B(-1, 1)$ is always $4$,then the equation for the locus of $P$ is

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