For any point P(x, y) on the hyperbola,

`text(distance)\ PB=sqrt(x^2+(y+2)^2`

`text(distance)\ PA=sqrt(x^2+(y-2)^2`

Since PBPA = 2 in our example, then:

`sqrt(x^2+(y+2)^2)-sqrt(x^2+(y-2))=2`

Rearrange:

`sqrt(x^2+(y+2)^2)=sqrt(x^2+(y-2))+2`

Square both sides:

`x^2+(y+2)^2` `=[x^2+(y-2)^2]` `+4sqrt(x^2+(y-2)^2)+4`

Expand brackets and simplify:

`2y-1=sqrt(x^2+(y-2)^2`

Square both sides again:

4y2 − 4y + 1 = x2 + y2 − 4y + 4

Simplifying gives the equation of our hyperbola:

`y^2-x^2/3=1`

The asymptotes (the red dotted boundary lines for the curve) are obtained by setting the above equation equal to `0`, rather than `1`.

`y^2-x^2/3=0`

This gives us the 2 lines:

`y=-x/sqrt3`, and   `y=x/sqrt3`

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