Question

$P(-3,6)$ is a given point and Q is a point on the locus $9x^2-16y^2=144$. Find the equation of the locus of midpoint of PQ.

Answer 1

$P(-3,6)$ let Q be $(x,y)$ lies on $9x^2-16y^2=144$

let mid-point be $(h,k)=(\frac{-3+x}{2},\frac{6+y}{2})$

=>$(x,y)=(2h+3,2k-6)$

$(x,y)$ lies on $9x^2-16y^2=144$

$=>9(2h+3{)}^2-16(2k-6{)}^2=144$

For locus replace $h\stackrel{}{\to }x$ ,$k\stackrel{}{\to }y=>\frac{(2x+3{)}^2}{16}-\frac{(2y+6{)}^2}{9}=1$