# equation of the locus of a point.

• Dec 23rd 2010, 08:00 PM
frankinaround
equation of the locus of a point.
I need help because I got stuck on this problem :

Find the equation of the locus of a point P = (x,y) that moves in accordance with each of the following conditions, and sketch the graphs:

a. The sum of the squares of the distances from P to the points (a,0) and (-a,0) is 4b^2, where b >= a/sqrt of 2 > 0

b. the distance of P from the point (8,0) is twice its distance from the point (0,4).

what ive done so far :

I think B is a line between the points (0,4) and (8,0) with a slope of 2/1. 2/1 is the negative reciprical of the line that goes threw (0,4) and (8,0)

So y=2X+B is the line I get for P according to the second guideline (b.).

Y and x of the point P must both work out to this formula so I should be able to substitute y for 2x+B in the equation I get from Part a of this problem. :
(x-a)^2+y^2+(x+a)^2+y^2=4B^2 =
(x-a)^2+(2x+B)^2+(x+a)^2+(2x+B)^2=4b^2 =
10x^2+2a^2+8xb-2b^2 = 0

Now I have the equation of some point I guess. But I dont know how to solve it from here. All I can think to do is simplify it all by dividing by 2. but then I get stuck. Someone please help.
• Dec 23rd 2010, 08:57 PM
mr fantastic
Quote:

Originally Posted by frankinaround
I need help because I got stuck on this problem :

Find the equation of the locus of a point P = (x,y) that moves in accordance with each of the following conditions, and sketch the graphs:

a. The sum of the squares of the distances from P to the points (a,0) and (-a,0) is 4b^2, where b >= a/sqrt of 2 > 0

b. the distance of P from the point (8,0) is twice its distance from the point (0,4).

what ive done so far :

I think B is a line between the points (0,4) and (8,0) with a slope of 2/1. 2/1 is the negative reciprical of the line that goes threw (0,4) and (8,0)

So y=2X+B is the line I get for P according to the second guideline (b.).

Y and x of the point P must both work out to this formula so I should be able to substitute y for 2x+B in the equation I get from Part a of this problem. :
(x-a)^2+y^2+(x+a)^2+y^2=4b^2
[snip]

Expand the left hand side. Simplify. Note the given condition on b.

As for the second locus: (x - 8)^2 + y^2 = 2(x^2 + (y - 4))^2.
Expand, simplify etc.
• Dec 27th 2010, 10:35 AM
earboth
Quote:

Originally Posted by frankinaround
I need help ...

b. the distance of P from the point (8,0) is twice its distance from the point (0,4).

1. The distance from P to A(8, 0) is:

$\displaystyle d_A=\sqrt{(x-8)^2+y^2}$

The distance from P to B(0,4) is:

$\displaystyle d_B=\sqrt{x^2+(y-4)^2}$

2. According to the question you'll get:

$\displaystyle \sqrt{(x-8)^2+y^2} = 2 \cdot \sqrt{x^2+(y-4)^2}~\implies~(x-8)^2+y^2=4(x^2+(y-4)^2)$

3. Expand the brackets, collect like terms and afterwards complete the squares.

4. For further informations have a look here: Circles of Apollonius - Wikipedia, the free encyclopedia