We see that a
D(x,y) is on the hyperbola
<=>
|D,F| - |D,F'| = (+ or -) 2a
<=>
_______________ _______________
| 2 2 | 2 2
\| (x - c) + y - \| (x + c) + y = (+ or -) 2a
Squaring <=>
________________________________
2 2 2 2 | 2 2 2 2 2
(x - c) + y + (x + c) + y - 2 \| ((x + c) + y ) ((x - c) + y ) = 4a
<=>
...
<=>
________________________________
| 2 2 2 2
\| ((x + c) + y ) ((x - c) + y ) = x2 + y2 + c2 - 2 a2
<=>
_______________________________________________
| 2 2 2 2 2 2
\| (x + c + y + 2 c x) (x + c + y - 2 c x) =(x2 + y2 + c2 )- 2 a2 (*)
Squaring =>
(x2 + y2 + c2 )2 - 4 x2 c = 4 a4 - 4 a2 (x2 + y2 + c2 ) + (x2 + y2 + c2 )2 (**)
<=>
- 4 x2 c = 4 a4 - 4 a2 (x2 + y2 + c2 )
<=>
...
<=>
(c2 - a2 ) x2 - a2 y2 = a2 (c2 - a2 )
Since a < c , we can say c2 - a2 = b2
<=>
b2 x2 - a2 y2 = a2 b2
<=>
x2 y2
-- - -- = 1 (***)
a2 b2
This is the equation of the hyperbola.
But we don't know if (*) and (**) are equivalent because we
don't know if the right
side of (*) is a positive value.
From above we know that if |D,F| - |D,F'| = 2a then (***) holds.
To prove the reverse, it is sufficient to show that if (***) holds,
then the right side of (*) is a positive value.
Well, from (***) we have
x2
-- > or = 1 => x2 > or = a2
a2
but also c2 > a2
-------------------
x2 + c2 > 2 a2
=> x2 + y2 + c2 > 2 a2
then the right side of (*) is a positive value.
Now abs(|D,F| - |D,F'|) = 2a and (***) are equivalent.
The intersection points of the hyperbola with the x-axis are A'(-a,0) and
A(a,0). These are the vertices on the x-axis.
The segment [A',A] is called major axis of the ellipse.
The segments [D,F'] and [D,F] are the focal radii through point D.
Orthogonal hyperbola as a special hyperbola
If a = b the equation becomes
x2 - y2 = a
This hyperbola is called an orthogonal hyperbola.
Parametric equations of the hyperbola
Take in a plane two lines l and m with resp. equations
x = a sec(t) (1)
y = b tan(t) (2)
The real number t is the parameter.
We know, from the theory of 'Elimination of parameters', that the
intersection points of the two associated lines constitute a curve.
To obtain the equation of that curve, we eliminate the parameter t
from the two equations. This means that we search for the condition
such that (1) and (2) has a solution for t.
The simultaneous equations (1) and (2) are equivalent to
a
x = ------
cos(t)
b sin(t)
y = -------
cos(t)
or to
a
cos(t) = -
x
a y
sin(t) = ---
b x
This system has a solution for t if and only if
sin2 (t) + cos2 (t) = 1
<=>
a 2 a y 2
(-) + (---) = 1
x b x
<=>
a2 b2 + a2 y2 = b2 x2
<=>
x2 y2
-- - --- = 1
a2 b2
Hence, the two associated lines constitute a curve and that curve
is the hyperbola.
We say that (1) and (2) are parametric equations of the hyperbola.
The point
D(a sec(t) , b tan(t))
is on the hyperbola for each t-value and with each point of the hyperbola
corresponds a t-value.
The hyperbola and to functions
x2 y2
-- - --- = 1
a2 b2
<=>
b2 (x2 - a2 )
y2 = -----------------
a2
<=>
_________ _________
b | 2 2 b | 2 2
y = - \| x - a or y = - - \| x - a
a a
We see that the hyperbola defines two functions.
The graph of the first one is above the x-as and the second graph is
the reflection of the first graph with respect to the x-axis.
If we calculate the asymptotes a and a', we find
b b
y = - x and y = - - x
a a
Tangent line in a point D of a hyperbola
In the same way as for the ellipse you'll find the equation of
the tangent line in point D(xo,yo)
xo x yo y
---- - ---- = 1
a2 b2
It is the bisector t of the lines DF and DF'.
Properties
It is easy to prove, with previous figure, that:
In the same way as for the ellipse, it is easy to prove that:
Topics and Problems
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