Let A=(2,0) and B=(0,-2). Let P be any point | vedclass

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(B) Let $P(x, y)$ be the point. Given $PA + PB = 4$. $\sqrt{(x-2)^2 + (y-0)^2} + \sqrt{(x-0)^2 + (y+2)^2} = 4$ $\sqrt{(x-2)^2 + y^2} = 4 - \sqrt{x^2 +

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$3x^2-2xy+3y^2-8x+8y=0$

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(B) Let $P(x, y)$ be the point. Given $PA + PB = 4$. $\sqrt{(x-2)^2 + (y-0)^2} + \sqrt{(x-0)^2 + (y+2)^2} = 4$ $\sqrt{(x-2)^2 + y^2} = 4 - \sqrt{x^2 + (y+2)^2}$ Squaring both sides: $(x-2)^2 + y^2 = 16 + x^2 + (y+2)^2 - 8\sqrt{x^2 + (y+2)^2}$ $x^2 - 4x + 4 + y^2 = 16 + x^2 + y^2 + 4y + 4 - 8\sqrt{x^2 + (y+2)^2}$ $-4x - 4y - 16 = -8\sqrt{x^2 + (y+2)^2}$ $x + y + 4 = 2\sqrt{x^2 + (y+2)^2}$ Squaring again: $(x + y + 4)^2 = 4(x^2 + y^2 + 4y + 4)$ $x^2 + y^2 + 16 + 2xy + 8x + 8y = 4x^2 + 4y^2 + 16y + 16$ $3x^2 - 2xy + 3y^2 - 8x + 8y = 0$.

Let $A=(2,0)$ and $B=(0,-2)$. Let $P$ be any point such that the sum of the distances of $P$ from $A$ and $B$ is $4$. Then the equation of the locus of the point $P$ is

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