Q.Show that the set of all points such that the difference of their distances from $(4, 0)$ and $(-4, 0)$ is always equal to 2 represent a hyperbola.

The key idea is the Locus of a Point, specifically the geometric definition of a hyperbola.

1. Let $P(x, y)$ be any point in the given set. The two fixed points are $F_1(4, 0)$ and $F_2(-4, 0)$.
2. The problem states that the difference of the distances from $P$ to $F_1$ and $F_2$ is always equal to 2. This can be written as $|PF_1 - PF_2| = 2$.
3. By definition, a hyperbola is the locus of all points such that the absolute difference of their distances from two fixed points (called foci) is a constant. …

The core idea here is understanding what a "locus of points" means and then recalling the geometric definition of a hyperbola.

A locus of points is simply the set of all points that satisfy a given geometric condition. For example, the locus of points equidistant from a single point is a circle.

The problem describes a very specific condition: the difference of distances from two fixed points is constant. This is precisely the definition of a hyperbola.

In this problem:

• The two fixed points (foci) are given as $F_1(4, 0)$ and $F_2(-4, 0)$.
• The constant difference of distances is given as 2. So, $2a = 2$, which implies $a = 1$.
• The distance between the foci is $2c$. Here, $2c = 4 - (-4) = 8$, so $c = 4$.
• For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We will use this to find $b^2$ after deriving the equation.

Our goal is to take the given condition, express it mathematically using the distance formula, and then simplify it algebraically to arrive at the standard equation of a hyperbola.

1. Set up the equation based on the given condition. Let $P(x, y)$ be any point in the locus. The two fixed points are $F_1(4, 0)$ and $F_2(-4, 0)$. The condition is that the difference of the distances from $P$ to $F_1$ and $F_2$ is 2. We must use the absolute difference to account for points where $PF_1 > PF_2$ or $PF_2 > PF_1$.

Using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

Substituting these into the condition:

This implies:

2. Isolate one square root and square both sides. To eliminate the square roots, we need to isolate one of them and square the equation. Let's move the second square root term to the right side:

Now, square both sides:

3. Simplify and isolate the remaining square root. Notice that $x^2$, $16$, and $y^2$ appear on both sides. We can cancel them out:

Now, gather all non-square root terms on one side:

Divide the entire equation by 4 to simplify:

> [!WARNING]
> When squaring an equation like $A = \pm B$, it becomes $A^2 = B^2$. The $\pm$ sign disappears because $(\pm B)^2 = B^2$. Be careful not to reintroduce it or make sign errors.

4. Square both sides again to eliminate the last square root. …