Question

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

Answer 1

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

Hence

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

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

Squaring 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

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

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

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

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

Squaring 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

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

This is the equation of hyperbola