Show that the set all points such that the difference of their distances from `(4,0)a n d(-4,0)` is always equal to `2`represents a hyperbola.

Explanation

Let the points be `P (x,y).`
`therefore` Distances of P from `(4,0)` `sqrt((x+4)^(2))+y^^(2))` …(i)
Now, `sqrt((x+4)^(2)+y^(2))-sqrt((x-4)^(2)+y^(2))=2`
`sqrt((x+4)^(2)+y^(2))=2+sqrt((x-4)^(2)+y^(2))`
On squaring both sides, we get
`x^(2)+8x+16+y^(2)=4+x^(2)-8x+16+y^(2)+4sqrt((x-4)^(2)+y^(2))`
`rArr16x-4=4sqrt((x-4)^(2)+y^(2))`
`rArr4(4x-1)=4sqrt((x-4)^(2)+y^(2))`
`rArr 16x^(2)-8x+1=x^(2)= 16-8x+y^(2)`
`rArr 15x^(2)-y^(2)=15` which is parabola.