Question

$A(5,0)$ and$B(-5,0)$ are two given points variable point $P$ such that $PA-PB=6$. Show that the equation of locus of P is $\frac{x^2}{a}-\frac{y^2}{b}=1$.Find $a+b$

Answer 1

Distance between two points $=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$PA=\sqrt{(x-5{)}^2+(y{)}^2}$

$PB=\sqrt{(x+5{)}^2+(y{)}^2}$

$PA-PB=6$

$PA=6+PB$

$(PA{)}^2=36+(PB{)}^2+12PB$

$(x-5{)}^2+y^2=36+(x+5{)}^2+y^2+12\sqrt{(x+5{)}^2+y^2}$

$x^2+25-10x=36+x^2+25+10x+12\sqrt{(x+5{)}^2+y^2}$

$12\sqrt{(x+5{)}^2+y^2}=20x+36$

$144(x^2+25+10x+y^2)=400x^2+1296+1440x$

$144(x{)}^2+3600+1440x+144y^2=400x^2+1296+1440x$

$256x^2-144y^2=2304$

$64x^2-36y^2=576$

$16x^2-9y^2=144$

$\frac{x^2}{9}-\frac{y^2}{16}=1$

$a+b=9+16=25$