Question: Locus & Parabola (1 Viewer)

[Fortify]

[ANSWERED] Question: Locus & Parabola

I'm stuck on this question:

Point P moves so that (PA)^2 + (PB)^2 = 12 where A = (-2,-5) and B = (1,3). Find the equation of the locus of P

 

[Arowana21]

Let P have points at (x,y)
I would assume that u would know the distance formula, so just use that, but since u r squaring it, then u do not need the square roots.
From then on, just simplify. and then make y in terms of x and then u should have ur locus.
if u dont know what distance formula is, it is [square root (x + x1)^2 - (y+y1)^2]

 

[Aplus]

Let P by point (x,y)
With the use of the distance formula:
(-2-x)² + (-5-y)² + (3-x)² + (1-y)² = 12
4 + 4x + x² + 25 + 10y + y² + 9 - 6x + x² + 1 - 2y + y² = 12
2x² - 2x +2y² + 10y = -26

Factorise:
2(x² - x) + 2(y² + 5y) = -26

Complete the square:
2(x-1)² + 2(y +5/2)² = -26 + 1 +25/4
2(x-1)² + 2(y +5/2)² = -75/4

Multiply both sides by 1/2:
(x-1)² + (y +5/2)² = -75/8

 

[Fortify]

Let P by point (x,y)
With the use of the distance formula:
(-2-x)² + (-5-y)² + (3-x)² + (1-y)² = 12
4 + 4x + x² + 25 + 10y + y² + 9 - 6x + x² + 1 - 2y + y² = 12
2x² - 2x +2y² + 10y = -26

Factorise:
2(x² - x) + 2(y² + 5y) = -26

Complete the square:
2(x-1)² + 2(y +5/2)² = -26 + 1 +25/4
2(x-1)² + 2(y +5/2)² = -75/4

Multiply both sides by 1/2:
(x-1)² + (y +5/2)² = -75/8

You got it wrong

 

[Just.Snaz]

EDIT: ok BIG edit.. i wrote the question wrong.. lol i really shouldn't do maths when I've just woken up...

 

[Just.Snaz]

okay.. 3rd time lucky maybe: is THIS the answer?
(x + 1/2)^2 + (y - 4)^2 = 11/4

 

[Fortify]

No, dw i solved it, it was actually really easy >_>;;

Thanks Arowana21 and Just.Snaz for his/her attempt.

 

[jellybelly59]

umm... here's another locus question just that i cbf to make a new thread umm... Find the eqn. of the locus of a pt that moves so that its distance fromt he line 3x + 4y + 5 = 0 is always 4 units

 

[lolokay]

umm... here's another locus question just that i cbf to make a new thread umm... Find the eqn. of the locus of a pt that moves so that its distance fromt he line 3x + 4y + 5 = 0 is always 4 units

3x+4y+5/sqrt.3^2+4^2 = 4
3x + 4y + 5 = 20
3x + 4y - 15 = 0

 

[jellybelly59]

ty lolokay

 

[tacogym27101990]

theres actually two lines that satisfy the equation

3x+4y-15=0 and 3y+4y+25=0

because when u have sqr(25) both 5 and negative five satisfy it so you have to account for two cases