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 PCMy 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