DERIVATION OF STANDARD EQUATION FOR HYPERBOLA
FROM THE LOCUS DEFINITION OF A HYPERBOLA
Left Diagram:
The difference of the distances from the point P to the foci is constant:
( x − (−c))2 + ( y − 0) 2 − ( x − c) 2 + ( y − 0) 2 = constant
Right Diagram:
In the case where the point on the hyperbola is a vertex V, we see that the difference of
the distances is 2a
( x − (−c))2 + ( y − 0) 2 − ( x − c) 2 + ( y − 0) 2 = 2a
This formula can be manipulated algebraically to rearrange it into
the standard form of the equation of a hyperbola
(see next page).
( x − (−c))2 + ( y − 0) 2 − ( x − c) 2 + ( y − 0) 2 = 2a
( x + c) 2 + y 2 − ( x − c) 2 + y 2 = 2a
( x + c) 2 + y 2 = 2a + ( x − c) 2 + y 2
(
( x + c) 2 + y 2 = 2a + ( x − c) 2 + y 2
(moved one radical to the other side)
)
2
(squared both sides)
x 2 + 2cx + c 2 + y 2 = 4a 2 + 4a ( x − c) 2 + y 2 + ( x − c) 2 + y 2
x 2 + 2cx + c 2 + y 2 = 4a 2 + 4a ( x − c) 2 + y 2 + x 2 − 2cx + c 2 + y 2
2cx = 4a 2 + 4a ( x − c) 2 + y 2 − 2cx
4cx − 4a 2 = 4a ( x − c) 2 + y 2
cx − a 2 = a ( x − c) 2 + y 2
(cx − a )
2 2
(
(finally finished simplifying!!!)
= a 2 ( x − c) 2 + y 2
)
(squared both sides, again)
c 2 x 2 − 2a 2 c 2 x 2 + a 4 = a 2 x 2 − 2a 2 c 2 x 2 + a 2 c 2 + a 2 y 2
c 2 x 2 + a 4 = a 2 x 2 + a 2c 2 + a 2 y 2
a 4 − a 2c 2 = a 2 x 2 − c 2 x 2 + a 2 y 2
(
a (c
) (
− a ) = (c
)
− a )x
a2 a2 − c2 = a2 − c2 x2 + a2 y2
2
2
2
2
2
2
− a2 y2
almost finished simplifying
multiplied by − 1, because c > a, so c2 > a2
now both sides are positive
Define b as the positive number such that b = c − a
2
a 2b 2 = b 2 x 2 − a 2 y 2
a 2b 2 b 2 x 2 a 2 y 2
=
−
a 2b 2 a 2b 2 a 2b 2
x2 y2
Finally we simplify to 1 = 2 − 2
a
b
2
2
(substituted b2 )
(divide both sides to get 1 on the constant side )
, standard form of equation of a hyperbola.
NOTE: We're not really adequately finished yet. To be thoroughly rigorous in proving the facts
we use about hyperbolas, we would still need to prove that the equations of the asymptotes are
( a) x . Proving that requires understanding limits, which are studied in calculus, so we
y=±b
can't prove it at our level of mathematics in a precalculus class.
The following proof uses limits, so requires Calculus 1A.
It is beyond the scope of Math 43 Precalculus.
You are not expected to understand this yet in Math 43.
x2 y2
b
Proof that the line y = x is the asymptote of the hyperbola 2 − 2 = 1
a
b
a
in the first quadrant.
The other quadrants follow by symmetry.
x2
y2
−1 = 2
a2
b
2

2 x
b  2 − 1 = y 2
a

 x2 
y = b  2 − 1 equation of hyperbola in first quadrant, with y as a function of x
a

2
We claim that y =
b
x is the equation of the asymptote in the first quadrant.
a
2

 b 
2 x


b
−
1
lim
 a 2  − a x 
x→∞ 




= lim
x→∞ 
a
(x
2
b
= lim
x→∞ 
a
(
2
b
)

− a2 − x 

) (( xx

x −a −x 

2
(
)
+ x)
2
− a2 + x
2
− a2
)
b x2 − a2 − x2 

2
2
a x − a + x 

= lim
x →∞
b

a2
lim


= x→∞
2
2
a
x
−
a
+
x




 = ab (0 ) = 0
2
2
 x −a + x

= ab lim
x →∞
1
 x2 
b
Therefore the curve y = b  2 − 1 approaches the line y = x as x → ∞
a
a

2