# Thread: Equation Of A Locus

1. ## Equation Of A Locus

,
I'm having trouble dealing with the following question. I need your help.

Q) Find the equation of the locus of a point P whose distance from (-1,1) is equal to thrice it's distance from the Y-axis.

The answer is reported as $\displaystyle 8x^2 - y^2 -2x +2y -2 = 0$, which i failed to get.

2. ## Equation of a locus

Hello saberteeth
Originally Posted by saberteeth
,
I'm having trouble dealing with the following question. I need your help.

Q) Find the equation of the locus of a point P whose distance from (-1,1) is equal to thrice it's distance from the Y-axis.

The answer is reported as $\displaystyle 8x^2 - y^2 -2x +2y -2 = 0$, which i failed to get.
Suppose the point P is $\displaystyle (x,y)$. Then its distance from $\displaystyle (-1,1)$ is $\displaystyle \sqrt{(x+1)^2+(y-1)^2}$, and its distance from the y-axis is simply $\displaystyle x$. So its locus has equation:

$\displaystyle \sqrt{(x+1)^2+(y-1)^2} = 3x$

i.e. $\displaystyle (x+1)^2 +(y-1)^2 = 9x^2$ (Did you forget to square the 3?)

i.e. $\displaystyle x^2+2x+1+y^2-2y+1 = 9x^2$

i.e. $\displaystyle 8x^2 -y^2 -2x+2y -2 =0$

3. Originally Posted by Grandad
Hello saberteethSuppose the point P is $\displaystyle (x,y)$. Then its distance from $\displaystyle (-1,1)$ is $\displaystyle \sqrt{(x+1)^2+(y-1)^2}$, and its distance from the y-axis is simply $\displaystyle x$. So its locus has equation:

$\displaystyle \sqrt{(x+1)^2+(y-1)^2} = 3x$

i.e. $\displaystyle (x+1)^2 +(y-1)^2 = 9x^2$ (Did you forget to square the 3?)

i.e. $\displaystyle x^2+2x+1+y^2-2y+1 = 9x^2$

i.e. $\displaystyle 8x^2 -y^2 -2x+2y -2 =0$