It is given that the difference of the distance between the points $(4,0)$ and $(-4,0)$ = 2

Hence $\sqrt{(x-4)^2+(y-0)^2}-\sqrt{(x+4)^2+(y-0)^2}=2$

$\Rightarrow \sqrt{(x-4)^2+y^2}=2+\sqrt{(x+4)^2+y^2}$

Squaring on both sides we get,

$(x-4)^2+y^2=4+(x+4)^2+y^2+4\sqrt{(x+4)^2+y^2}$

On expanding we get,

$x^2-8x+16+y^2=4+x^2+8x+16+y^2+4\sqrt{(x+4)^2+y^2}$

$-16x-4=4\sqrt{(x+4)^2+y^2}$

$-4(4x+1)=4\sqrt{(x+4)^2+y^2}$

$\Rightarrow -(4x+1)=\sqrt{(x+4)^2+y^2}$

Squaring on both sides we get,

$16x^2+8x+1=x^2+8x+16+y^2$

$15x^2-y^2=15$

Dividing throughout by 15 we get,

$\large\frac{x^2}{1}-\frac{y^2}{15}$$=1$

This is clearly the equation of a hyperbola.

Hence proved.