Hyperbole formula (1 Viewer)

[sifANDREW]

Hey i was just wondering how you would derive the hyperbola formula. This is how i derived the ellipse formula:
Ellipse with major axis (x) 2a and minor axis (y) 2b, center (0,0)
point A(a,0) is at extremity of major axis, B(0,b) is extremity at minor axis
Point S (k,0) and S'(-k,0) where 0<k<a [focus] and line L (x=n) and L'(x=-n) [directrix]
Point P(x,y) lies on curve


SA/AL = e
SA=eAL
a-k = e(n-a) [eqn1]

S'A/AL' = e
S'A=EAL'
a+k = e(n+a) [eqn2]

SOLVING SIMULTANEOUSLY
k = ae
n = a/e
hence focus (ae,0) and directrix (x=a/e)

PS/PM = e [PM is perpendicular distance from P to x=ae directrix]
PS=ePM
using distance formula etc collect like terms and you get.....

x^2(1-e^2) + y^2 = a^2(1-e^2) [eqn3]

THIS IS THE PART WHERE IM CONFUSED
i sub in the point B (0,b) since it lies on the curve, hence
b^2 = a^2(1-e^2)[eqn4]

and hence i divide both sides of [eqn3] by b^2 and get

x^2/a^2 + y^2/b^2 = 1

which is the ellipse formula.

you do exact same thing for the hyperbola formula HOWEVER i cannot derive eqn4 for hyperbola since i do not know the "extreme point B" for the hyperbola. How do i derive the equation b^2=a^2(e^2-1) for the hyperbola?

and if i done anything wrong please tell me as well.

Thanks for the help!

edit: unless i actually start with the hyperbola formula and sub in y=b.... which isnt really deriving i guess lol.
edit: yes LOL hyperbole is an english technique my bad LMAOS =p

 

[MetroMattums]

You don't derive b^2 = a^2 (e^2 - 1), you postulate it :/

 

[hscishard]

Isn't hyperbole an english technique?

 

[sifANDREW]

Hey uhm...can you explain to me how you actually think of postulating it? Because i dont understand how you can show that 'b' is the extreme point on the hyperbola. So yea can you just explain how you postulate it and it's thought processes thanks lol.

 

[MetroMattums]

Well, there is no extreme point denoted by 'b' on the hyperbola, it only applies to circles and ellipses :/

b is basically a function of the eccentricity of the hyperbola - as e > 1, it can reach infinity so your assertion that b is the extreme point of the hyperbola is kinda irrelevant..

Oh. And for the ellipse. When you sub in (0, b) or whatever extreme point you have that is not logically correct - you have to state that b^2 = a^2(1 - e^2) FIRST, then you can state that the extreme point is (0, b)

Otherwise you're just pulling a constant b out of nowhere. You have to state where it comes from, hence the postulation.

And I dislike the term 'extreme point' <_<

 

[sifANDREW]

Sorry i still dont really understand the whole postulating thing. How can someone just think of let b^2 = a^2(1-e^2) or let b^2 = a^2(e^2-1)
Or is that the whole point of a postulate and then by using values of "e" you equate b?

Since P(x,y) is any point of the ellipse curve, can't i sub in the point B(0,b) since it also lies on the curve?

sorry i couldnt think of another term to use @.@ i understand what your saying about the term extreme point though.

and im not really like full on bright but yea im trying so help is appreciated. Forgive my ignorance.

Thanks for the help!

 

[MetroMattums]

Yeah it does seem kinda random.

However, it's essentially a simplification of the term a^2(e^2 - 1), nothing else. Even without b you would still have the same graph, except you'd notate b as a√(e^2 - 1)

Regarding subbing in B(0,b): at this point in your proof you give no indication of the existence of b - it's logically wrong to simply introduce a constant without prior postulation of b^2 = blah blah

TL;DR = b is just what mathematicians use to make their equations nice and neat

EDIT: Your derivation should start from PS = ePM, having defined a fixed point (the focus) and a line (the directrix) anyway

 

[sifANDREW]

Yep got it. Thanks for clarification!