Find the locus of point $P$ such that $AP^2 - BP^2 = 1$, where $A$ and $B$ are two fixed points and $AB=2$.

1. Problem Description

Find the locus of point $P$ such that $AP^2 - BP^2 = 1$, where $A$ and $B$ are two fixed points and $AB=2$.

2. Solution Steps

Let $P(x, y)$, $A(1, 0)$ and $B(-1, 0)$. We are given that $AB = 2$ and $AP^2 - BP^2 = 1$.

$AP^2 = (x-1)^2 + (y-0)^2 = (x-1)^2 + y^2 = x^2 - 2x + 1 + y^2$

$BP^2 = (x+1)^2 + (y-0)^2 = (x+1)^2 + y^2 = x^2 + 2x + 1 + y^2$

Now we have $AP^2 - BP^2 = (x^2 - 2x + 1 + y^2) - (x^2 + 2x + 1 + y^2) = 1$.

Simplifying the expression:

$x^2 - 2x + 1 + y^2 - x^2 - 2x - 1 - y^2 = 1$

$-4x = 1$

$x = -\frac{1}{4}$

The locus of $P$ is the vertical line $x = -\frac{1}{4}$.

3. Final Answer

$x = -\frac{1}{4}$