What is the equation of this curve with P given? The difference of the distance between P(x,y) and the points (-2,1) and (8,1) is 6.

1 Answer
May 13, 2018

#16x^2-9y^2-96x+18y-9=0#

Explanation:

let the point be #P(x,y)# (this will be general point).
Now according to question we have points
#A(-2,1)# and #B(8,1)#

also, the difference between the distances of the point is 6 units #i.e# #PA-PB=6#
or # PA=PB+6#
#=>PA^2=(PB+6)^2#
#=>PA^2=PB^2+12PB+36# #...(i)#
My PC

by distance formula we have
#PA=sqrt((x+2)^2+(y-1)^2)# and #PB=sqrt((x-8)^2+(y-1)^2#
#=>PA^2=(x+2)^2+(y-1)^2# and #PB^2=(x-8)^2+(y-1)^2#
putting these in #(i)#
we get
#(x+2)^2+cancel((y-1)^2)=(x-8)^2+cancel((y-1)^2)+12sqrt((x-8)^2+(y-1)^2)+36#

on solving we'll get
#16x^2-9y^2-96x+18y-9=0#
this represents a hyperbola,
graph{16x^2-9y^2-96x+18y-9=0 [-16.02, 16.02, -8.01, 8.01]}

this will be the equation of variable point #P(x,y)#
this is called the LOCUS of point #P#