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 *PB* − *PA* = 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:

4*y*^{2} − 4*y* + 1 = *x*^{2 }+* y*^{2} − 4*y* + 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`

Get the Daily Math Tweet!

IntMath on Twitter