# When to introduce absolute value in hyperbola expression?

1. Nov 27, 2015

### Odious Suspect

We begin with this definition of a hyperbola.

$$\left(\overline{F_1 P}-\overline{F_2 P}=2 a\right)\land a>0$$

Perform a few basic algebraic manipulations.

$$\sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a$$

$$\sqrt{(c+x)^2+y^2}=2 a+\sqrt{(x-c)^2+y^2}$$

$$(c+x)^2+y^2=4 a^2+4 a \sqrt{(x-c)^2+y^2}+(x-c)^2+y^2$$

Clearly both sides of this equation must be non-negative. Expand the squares.

$$c^2+2 c x+x^2+y^2=4 a^2+4 a \sqrt{(x-c)^2+y^2}+c^2-2 c x+x^2+y^2$$

Again, it appears that both sides must be non-negative. The equation appears to valid without an absolute value symbol.

Now we subtract away some non-negative terms.

$$2 c x=4 a^2+4 a \sqrt{(x-c)^2+y^2}-2 c x$$

If $$x<0$$ there are no real solutions. We can patch it up with absolute value symbols. But when should that restriction first be introduced?

$$c x-a^2=a \sqrt{(x-c)^2+y^2}$$
$$\mid \frac{c}{a} x-a \mid = \sqrt{(x-c)^2+y^2}$$

Last edited: Nov 27, 2015
2. Nov 27, 2015

### BvU

You introduced it when you gave your own definition of the hyperbola. Ususally we have $|2a|$ there.

3. Nov 27, 2015

### Odious Suspect

That certainly is universal. For example see for example http://mathworld.wolfram.com/Hyperbola.html. The definition I followed is found in Thomas's Classical Edition of Calculus with Analytic Geometry.

4. Nov 28, 2015

### BvU

Example: a = 1 c = 2. The point $(1,0)$ is on 'your' hyperbola, the point $(-1,0)$ is not ?

I must be overlooking something; can't even reproduce the step from wolfram (3) to (4)...

5. Nov 28, 2015

### Samy_A

This is not correct. It is the absolute value of the difference of the distances to the two foci that must be equal to 2a.
If c>0, here you implicitly assume that x>0, as you assume that the distance from (x,y) to (-c,0) is larger than the distance from (x,y) to (c,0).

6. Nov 28, 2015

### Odious Suspect

Distance is positive definite in Euclidean space.

7. Nov 28, 2015

### Samy_A

Sure, but difference of distances is not. A point belongs to the hyperbola if the absolute value of the difference of the distances to the two foci is equal to 2a.

Using your formula, for x>0, (x,y) is on the hyperbola if:
$\sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a$ $\ \ (1)$

For x<0, (x,y) is on the hyperbola if:
$\sqrt{(c-x)^2+y^2}-\sqrt{(x+c)^2+y^2}=2 a$ $\ \ (2)$

If you want one expression covering both cases, take the absolute value:
$|\sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}|=2 a$ $\ \ (3)$
$\frac{x²}{a²}-\frac{y²}{c²-a²}=1$ $\ \ (4)$
(1,0) satisfies equation $(1)$, (-1,0) satisfies equation $(2)$. Both satisfy equations $(3)$ and $(4)$.