# 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:
$|\sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}|=2 a$ $\ \ (3)$
Or, use the equation derived in your Mathworld.Wolfram link:
$\frac{x²}{a²}-\frac{y²}{c²-a²}=1$ $\ \ (4)$

@BvU gave an example in post 4, with a=1 and c=2.
(1,0) satisfies equation $(1)$, (-1,0) satisfies equation $(2)$. Both satisfy equations $(3)$ and $(4)$.

Last edited: Nov 28, 2015
8. Nov 29, 2015

### Odious Suspect

Indeed. I was going off my notes and recollection of Thomas. He actually introduced the absolute value signs in his derivation of the ellipse, and then said they were superfluous in that case. But his development of the hyperbola works your two cases in parallel. Taking the absolute value of the difference of the distances, as you did appears to accomplish the same thing.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook