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

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Aicharya · Accepted Answer

Let point $P(x, y)$, so $ \begin{aligned} & \sqrt{(x-1)^2+(y+1)^2}+\sqrt{(x+1)^2+(y-1)^2}=4 \ & \Rightarrow(x-1)^2+(y+1)^2+(x+1)^2+(y-1)^2 \ & +2 \sqrt{(x-1)^2+(y+1)^2} \sqrt{(x+1)^2+(y-1)^2}=16 \ & \Rightarrow 2\left(x^2+y^2+2ight)+2 \sqrt{(x-1)^2+(y+1)^2} \ & \sqrt{(x+1)^2+(y-1)^2}=16 \ & \Rightarrow \quad\left(x^2+y^2+2ight)+\sqrt{(x-1)^2+(y+1)^2} \ & \sqrt{(x+1)^2+(y-1)^2}=8 \ & \Rightarrow\left[(x-1)^2+(y+1)^2ight]\left[(x+1)^2+(y-1)^2ight] \ & =\left(6-x^2-y^2ight)^2 \ & \Rightarrow\left[x^2+y^2-2 x+2 y+2ight]\left[x^2+y^2+2 x-2 y+2ight] \ & =\left(x^2+y^2-6ight)^2 \ & \end{aligned} $ $\begin{aligned} & \Rightarrow\left(x^2+y^2+2^2-4(x-y)^2=\left(x^2+y^2-6ight)^2ight. \ & \Rightarrow \quad\left[\left(x^2+y^2+2ight)-\left(x^2+y^2-6ight)ight] \ & {\left[\left(x^2+y^2+2ight)+\left(x^2+y^2-6ight)ight]} \ & =4(x-y)^2 \ & \Rightarrow \quad 8\left(2 x^2+2 y^2-4ight)=4\left(x^2+y^2-2 x yight) \ & \Rightarrow \quad 4 x^2+4 y^2-8=x^2+y^2-2 x y \ & \Rightarrow \quad 3 x^2+2 x y+3 y^2=8 \ & \end{aligned}$

If a point $P$ moves such that the sum of the distances from $P$ to the point $A(1,-1)$ and $B(-1,1)$ is…

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