AP Calculus BC class summer preparation work help?? im stuck on this one problem...?

Question

The problem is: A curve is traced by a point P(x, y) which moves such that its distance from the point A(-1, 1) is three times its distance from the point B(2, -1). Determine the equation of the curve.

Ok so ive already figured out that one of the points of the curve is (1.25, -0.5), and the answer to the problem is that the equation of the cuve is 8x^2 - 38x + 8y^2 + 20y + 43 = 0 which is an equation of a circle. but how do i get from the little info that i have to the answer?? please help me

Answer 1

Just read the question and try making the equation.Frm the question,
PA=3PB
->PA^2=9PB^2
->(x+1)^2+(y-1)^2=9[(x-2)^2+(y+1)^2]
and then simplify the above.U shud get the equation.I dont think thr is any calculus involvd

Answer 2

let (x,y) is the point on the curve
3 sqr[ (x+1)^2 + (y-1)^2 ] = sqr[ (x-2)^2 + (y+1)^2] . . .squaring both sides
9[ x^2 + 2x +1 + y^2 -2y + 1 ] = x^2+4x +4 + y^2 +2y +1
8x^2 - 14x + 8 Y^2 - 20y = -13
x^2 - 14x/8 + Y^2 - 20y /8= -13/8
x^2 - 7x/4 + (7/8)^2 + Y^2 - 5y/2 + (5/4)^2 = -13/8+ (7/8)^2 + (5/4)^2
(x - 7/8)^2 +(y-5/4)^2 = 45/64 = (sqr(45)/8)^2 . . . equation of circle

It is a circle of center (7/8 , 5/4) . . . radius = sqr(45)/8

Answer 3

write the distances

sqrt[(x-(-1))^2+(y-1)^2]=3 * sqrt[(x-2))^2+(y+1)^2]

take square of both sides

x^2+2x+1+y^2-2y+1
=9(x^2-4x+4+y^2+2y+1)


8x^2 - 38x + 8y^2 + 20y + 43 = 0

Answer 4

you ought to take it as long because it does not shrink to rubble your first semester grades. by 2d semester, you should be in senior droop mode anyhow. by taking it and passing the AP examination, you ought to get college credit for it. that's what I did, yet ended up taking it anyhow back in college because of the fact it exchange into an elementary thank you to get a extreme grade. did no longer even ought to income for it.