Locus question (1 Viewer)

[chewy123]

Find expression for:

The point P(x,y) moves so that its distance PA from the point A(1,5) is always twice its distance PB from the point B(4,-1).

Ans: x^2 + y^2 -10x +6y +14 =0

If you have the jones & couchman book it is EX: 17.2 Q6

Unfortunately they only have examples for equidistant and I tried multiplying the other side by 2 and solve, but doesn't work

Well, thx to any helpers.

 

[ssglain]

This question is to be solved by exactly the same principle as when P is equidistant from A and B. The conditions have simply changed to PA = 2PB rather than PA = PB.

Spoiler

P(x, y) A(1, 5) B(4, -1)

PA² = (x - 1)² + (y - 5)² = x² - 2x + y² - 10y + 26
PB² = (x - 4)² + (y + 1)² = x² - 8x + y² + 2y +17

Now, PA = 2PB --> PA² = 4PB²
x² - 2x + y² - 10y + 26 = 4(x² - 8x + y² + 2y +17)
x² - 2x + y² - 10y + 26 = 4x² - 32x + 4y² + 8y + 68
.: 3x² - 30x - 3y² + 18y + 42 = 0
i.e. x² - 10x - y² + 6y + 14 = 0 is the equation of the locus.

 

[chewy123]

Thx, and a follow up question

What is the equation satisfied by all points whose distance from the line y=-2 is the same as their distance from th point (0,2)

I don't understand what they are asking at all

 

[independantz]

P(x,y) A( 0,2) B(x,-2)
PA=PB
sqrt(x)^2+(y-2)^2=sqrt(x-x)^2+(y+2)^2
square both sides
x^2+y^2-4y+4=y^2+4y+4
x^2=8y
I think that's how it's done.