When to introduce absolute value in hyperbola expression?

[Odious Suspect]

We begin with this definition of a hyperbola.

[tex]\left(\overline{F_1 P}-\overline{F_2 P}=2 a\right)\land a>0[/tex]

Perform a few basic algebraic manipulations.

[tex]\sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a[/tex]

[tex]\sqrt{(c+x)^2+y^2}=2 a+\sqrt{(x-c)^2+y^2}[/tex]

[tex](c+x)^2+y^2=4 a^2+4 a \sqrt{(x-c)^2+y^2}+(x-c)^2+y^2[/tex]

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

[tex]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[/tex]

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.

[tex]2 c x=4 a^2+4 a \sqrt{(x-c)^2+y^2}-2 c x[/tex]

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

[tex]c x-a^2=a \sqrt{(x-c)^2+y^2}[/tex]
[tex]\mid \frac{c}{a} x-a \mid = \sqrt{(x-c)^2+y^2}[/tex]

 

[BvU]

But when should that restriction first be introduced?

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

 

[Odious Suspect]

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

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.

 

[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)...

 

[Samy_A]

We begin with this definition of a hyperbola.

[tex]\left(\overline{F_1 P}-\overline{F_2 P}=2 a\right)\land a>0[/tex]

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.

Perform a few basic algebraic manipulations.

[tex]\sqrt{(c+x)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a[/tex]

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).

 

[Odious Suspect]

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).

Distance is positive definite in Euclidean space.

 

[Samy_A]

Distance is positive definite in Euclidean space.

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)##.

 

[Odious Suspect]

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)##.

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.