# Thread: How to find the equation of the locus of this point?

1. ## How to find the equation of the locus of this point?

The question: Find the equation of the locus of the point which is equidistant from the y-axis and the point (3,-1).

So I let the variable point be P(x,y)
for the y-axis I use A(0,y)
and I call the given point B(3,-1)

I then form the equation PA^2=PB^2 and solve.

I can't seem to find the correct answer. If anyone could point out what I'm doing wrong I'd really appreciate it

2. ## Re: How to find the equation of the locus of this point?

The locus of points is a parabola ...

focus, $(3,-1)$

directrix, $x=0$

vertex, $(h,k) = (1.5,-1)$

$(y-k)^2 = 4p(x-h)$, where $p$ is half the distance between the focus and directrix.

3. ## Re: How to find the equation of the locus of this point?

Originally Posted by Quoctopus
The question: Find the equation of the locus of the point which is equidistant from the y-axis and the point (3,-1).

So I let the variable point be P(x,y)
for the y-axis I use A(0,y)
and I call the given point B(3,-1)

I then form the equation PA^2=PB^2 and solve.

I can't seem to find the correct answer. If anyone could point out what I'm doing wrong I'd really appreciate it
It's impossible to point out what you are doing wrong when you don't tell us what you did! HOW did you "form the equation PA^2= PB^2"?