Question

Find the equation of locus of a point, the difference of whose distances from $(-5,0)$ and $(5,0)$ is $8$

Answer 1

Let the point be (x,y),
By using distance formula for coordinates,
$\left[\right(x+5{)}^2+(y-0{)}^2{]}^{\frac{1}{2}}-[(x-5{)}^2+(y-0{)}^2{]}^{\frac{1}{2}}=8$
$\Rightarrow [x^2+10x+25+y^2{]}^{\frac{1}{2}}=[x^2-10x+25+y^2{]}^{\frac{1}{2}}+8$
squaring on both sides,
$\Rightarrow x^2+10x+25+y^2=x^2-10x+25+y^2+64+16[x^2-10x+25+y^2{]}^{\frac{1}{2}}$
$\Rightarrow 5x-16=4[x^2-10x+25+y^2{]}^{\frac{1}{2}}$
Again squaring both sides,
$\Rightarrow 25x^2+256-160x=16x^2-160x+400+16y^2$
Therefore, locus is,
$9x^2-16y^2=144$.......(hyperbola)