[SOLVED] Finding equation of locus

**xwrathbringerx** · April 14th 2009, 01:30 AM

Hey guys

A pt P(x,y) moves so that its distance from (3,4) is proportional to its distance from (-1,2). Find the equation of the locus of P if the origin is a pt on the locus.

I seriously have no clue how to do this q. - i tried using direct variation i.e.
$(0-3)^2 + (0-4)^2 = k^2 (0+1)^2 + (0-2)^2$
...
$k = +/- 2$
then subbing k into $(x-3)^2 + (y-4)^2 = k^2 (x+1)^2 + (y-2)^2$

Naturally it didn't work =P

Help please?

Thanx a lot.

**xwrathbringerx** · April 14th 2009, 01:50 AM

Actually i fink i've solved it YAY ...sori...how do i delete this thread?

**mr fantastic** · April 14th 2009, 02:05 AM

Originally Posted by xwrathbringerx

Actually i fink i've solved it YAY ...sori...how do i delete this thread?

Why not post your solution and get an opinion of it ....

**xwrathbringerx** · April 14th 2009, 02:55 PM

Is this how u do it? Mayb it's wrong because the textbook's answer is actually $x^2 + y^2 + x - 3y = 0$ instead of $x^2 + y^2 + 4x - 3y = 0$

Let pt (3,4) be A and pt (-1,2) be B.
AP is proportional to BP.
So $AP^2$ is proportional to $BP^2$ .
Therefore, AP = k × BP
$(x-3)^2 + (x-4)^2 = k ( (x+1)^2 + (x-2)^2)$
Substitute in (0,0)
$(0-3)^2 + (0-4)^2 = k( (0+1)^2 + (0-2)^2)$
9 + 16 = k (1+4)
25 = 5k
Therefore, k = 5
Substitute k = 5 into the original equation.
$(x-3)^2 + (x-4)^2 = 5 ( (x+1)^2 + (x-2)^2)$
$x^2 - 6x + 9 + y^2 - 8y + 16 = 5(x^2 + 2x + 1 + y^2 - 4y + 4)$
$x^2 + y^2 - 6x - 8y + 25 = 5(x^2 + 2x + y^2 - 4y + 5)$
$= 5x^2 + 10x + 5y^2 - 20y + 25$
Hence, $x^2 + y^2 + 4x - 3y = 0$

**mr fantastic** · April 14th 2009, 03:16 PM

Originally Posted by xwrathbringerx

Is this how u do it? Mayb it's wrong because the textbook's answer is actually $x^2 + y^2 + x - 3y = 0$ instead of $x^2 + y^2 + 4x - 3y = 0$

Let pt (3,4) be A and pt (-1,2) be B.
AP is proportional to BP.
So $AP^2$ is proportional to $BP^2$ .
Therefore, AP = k × BP
$(x-3)^2 + (x-4)^2 = k ( (x+1)^2 + (x-2)^2)$
Substitute in (0,0)
$(0-3)^2 + (0-4)^2 = k( (0+1)^2 + (0-2)^2)$
9 + 16 = k (1+4)
25 = 5k
Therefore, k = 5
Substitute k = 5 into the original equation.
$(x-3)^2 + (x-4)^2 = 5 ( (x+1)^2 + (x-2)^2)$
$x^2 - 6x + 9 + y^2 - 8y + 16 = 5(x^2 + 2x + 1 + y^2 - 4y + 4)$
$x^2 + y^2 - 6x - 8y + 25 = 5(x^2 + 2x + y^2 - 4y + 5)$
$= 5x^2 + 10x + 5y^2 - 20y + 25$
Hence, $x^2 + y^2 + 4x - 3y = 0$

At a glance I don't see anything wrong with your answer (but there are a couple of typos in the earlier parts of the solution, x's where there should be y's).

The books answer would be correct if the second point was (1, 2) rather than (-1, 2) , so maybe there's a typo somewhere.

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