Locus Question (1 Viewer)

[frenzal_dude]

Find the locus of a point P which moves so that:

PA^2 - PB^2 = 5

I don't know how to work this out without first knowing the location of A and B.
Hope you guys can help!

 

[life92]

Basically you just need to allocate A and B positions.

So let
A = (x1,y1)
B = (x2,y2)
P is a variable so (x,y)

Therefore, using the distance formula

(PA)^2 = (x-x1)^2+(y-y1)^2

(PB)^2 = (x-x2)^2+(y-y1)^2

Then the difference between these equals 5

So
[(x-x1)^2+(y-y1)^2]-[(x-x2)^2+(y-y1)^2] = 5

(x-x1)^2+(y-y1)^2-(x-x2)^2-(y-y1)^2 = 5
Using difference of squares
[(x-x1)-(x-x2)][(x-x1)+(x-x2)] + [(y-y1)-(y-y2)][(y-y1)+(y-y2)] = 5
(x2-x1)[2x-(x2+x1)] + (y2-y1)[2y-(y2+y1)] = 5
2x(x2-x1) - [(x2)^2-(x1)^2] + 2y(y2-y1) - [(y2)^2-(y1)^2] = 5

Now, make y the subject.
2y(y2-y1) = - 2x(x2-x1) + [(x2)^2-(x1)^2] + [(y2)^2-(y1)^2]+5
y = {- 2x(x2-x1) + [(x2)^2-(x1)^2] + [(y2)^2-(y1)^2]+5 } / [y2-y1]
If we examine this equation very carefully, it's actually a straight line in the form y=mx+b since x1,x2,y1,y2 are all constants.

It looks really complicated but yeah o_o
Hrmm,,
It'd simplify quite nicely if there were values for A and B I'm pretty sure...

 

[frenzal_dude]

hey thanks for your help.

In the book (New Senior Mathematics: Two unit course for yr 11&12, Fitzpatrick, pg 224, question 12a) they've got the answer as: 8x + 8y - 3 = 0

How can they get this answer without giving us A and B coordinates?

 

[life92]

Hrm..
If it simplifies to that, there's probably values for A and B somewhere.
Maybe it relates to a previous question or something?
Sorry I can't check myself because I don't have the Fitzpatrick 2U book.

 

[frenzal_dude]

ahk you're right. In the next part of the question, it has a different problem and they give you A(2,3) and B(-2,-1), badly written question.
thanks anyway tho