DERIVATION OF STANDARD EQUATION FOR HYPERBOLA FROM THE ...

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 + y2 − (x − c)2 + y2 = 2a

(x + c)2 + y2 = 2a + (x − c)2 + y2 (moved one radical to the other side)
(squared both sides)
2

( )(x + c)2 + y2 = 2a + (x − c)2 + y2

x2 + 2cx + c2 + y 2 = 4a 2 + 4a (x − c)2 + y 2 + (x − c)2 + y 2

x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 + x2 − 2cx + c2 + y2

2cx = 4a2 + 4a (x − c)2 + y2 − 2cx

4cx − 4a2 = 4a (x − c)2 + y2

cx − a2 = a (x − c)2 + y2 (finally finished simplifying!!!)

( ) ( )cx − a2 2 = a2 (x − c)2 + y2 (squared both sides, again)

c2 x2 − 2a2c2 x2 + a4 = a2 x2 − 2a2c2 x2 + a2c2 + a2 y2

c2x2 + a4 = a2x2 + a2c2 + a2 y2

a4 − a2c2 = a2x2 − c2x2 + a2 y2

( ) ( )a2 a2 − c2 = a2 − c2 x2 + a2 y2 almost finished simplifying

( ) ( )a2 c2 − a2 = c2 − a2 x2 − a2 y2 multiplied by − 1, because c > a, so c2 > a2
now both sides are positive

Define b as the positive number such that b2 = c2 − a2

a2b2 = b2x2 − a2 y2 (substituted b2 )

a2b2 = b2x2 − a2 y2 (divide both sides to get 1 on the constant side )
a2b2 a2b2 a2b2

Finally we simplify to 1= x2 − y2 , standard form of equation of a hyperbola.
a2 b2

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

( )y = ± b a x . Proving that requires understanding limits, which are studied in calculus, so we

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.

Proof that the line y=bx is the asymptote of the hyperbola x2 − y2 =1
a a2 b2

in the first quadrant. The other quadrants follow by symmetry.

x2 −1 = y2
a2 b2

b 2  x 2 − 1 = y2
a 2

y= b2  x2 −1 equation of hyperbola in first quadrant, with y as a function of x
a2

b
We claim that y = a x is the equation of the asymptote in the first quadrant.

 b2  x2 −1 − b 
lim  a2 a x
x→∞  

( )= lim b x2 − a2 − x 
x→∞  a 

( ) ( )=lim
( )x→∞
b x2 − a2 − x  x2 −a2 + x
 a  x2 −a2 + x

( )= lim  b 
x→∞ a 
x2 − a2 − x2

x2 −a2 + x 

lim  b a2 
a 
= x→∞ x2 −a2 + x

 1 
lim  − a2 
= ab  x2 + x  = ab (0 ) = 0
x→∞

Therefore the curve y = b 2  x 2 −1 b
a 2 a x as x → ∞
approaches the line y =