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10.2 The hyperbola  (Page 2/13)

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Deriving the equation of an ellipse centered at the origin

Let   ( − c , 0 )   and   ( c , 0 )   be the foci    of a hyperbola centered at the origin. The hyperbola is the set of all points   ( x , y )   such that the difference of the distances from   ( x , y )   to the foci is constant. See [link] .

If   ( a , 0 )   is a vertex of the hyperbola, the distance from   ( − c , 0 )   to   ( a , 0 )   is   a − ( − c ) = a + c .   The distance from   ( c , 0 )   to   ( a , 0 )   is   c − a .   The sum of the distances from the foci to the vertex is

( a + c ) − ( c − a ) = 2 a

If   ( x , y )   is a point on the hyperbola, we can define the following variables:

d 2 = the distance from  ( − c , 0 )  to  ( x , y ) d 1 = the distance from  ( c , 0 )  to  ( x , y )

By definition of a hyperbola,   d 2 − d 1   is constant for any point   ( x , y )   on the hyperbola. We know that the difference of these distances is   2 a   for the vertex   ( a , 0 ) .   It follows that   d 2 − d 1 = 2 a   for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula . The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.

                                       d 2 − d 1 = ( x − ( − c ) ) 2 + ( y − 0 ) 2 − ( x − c ) 2 + ( y − 0 ) 2 = 2 a Distance Formula ( x + c ) 2 + y 2 − ( x − c ) 2 + y 2 = 2 a Simplify expressions .                             ( x + c ) 2 + y 2 = 2 a + ( x − c ) 2 + y 2 Move radical to opposite side .                               ( x + c ) 2 + y 2 = ( 2 a + ( x − c ) 2 + y 2 ) 2 Square both sides .                      x 2 + 2 c x + c 2 + y 2 = 4 a 2 + 4 a ( x − c ) 2 + y 2 + ( x − c ) 2 + y 2 Expand the squares .                      x 2 + 2 c x + c 2 + y 2 = 4 a 2 + 4 a ( x − c ) 2 + y 2 + x 2 − 2 c x + c 2 + y 2 Expand remaining square .                                               2 c x = 4 a 2 + 4 a ( x − c ) 2 + y 2 − 2 c x Combine like terms .                                    4 c x − 4 a 2 = 4 a ( x − c ) 2 + y 2 Isolate the radical .                                        c x − a 2 = a ( x − c ) 2 + y 2 Divide by 4 .                                    ( c x − a 2 ) 2 = a 2 [ ( x − c ) 2 + y 2 ] 2 Square both sides .                      c 2 x 2 − 2 a 2 c x + a 4 = a 2 ( x 2 − 2 c x + c 2 + y 2 ) Expand the squares .                     c 2 x 2 − 2 a 2 c x + a 4 = a 2 x 2 − 2 a 2 c x + a 2 c 2 + a 2 y 2 Distribute  a 2 .                                    a 4 + c 2 x 2 = a 2 x 2 + a 2 c 2 + a 2 y 2 Combine like terms .                   c 2 x 2 − a 2 x 2 − a 2 y 2 = a 2 c 2 − a 4 Rearrange terms .                     x 2 ( c 2 − a 2 ) − a 2 y 2 = a 2 ( c 2 − a 2 ) Factor common terms .                               x 2 b 2 − a 2 y 2 = a 2 b 2 Set  b 2 = c 2 − a 2 .                              x 2 b 2 a 2 b 2 − a 2 y 2 a 2 b 2 = a 2 b 2 a 2 b 2 Divide both sides by  a 2 b 2                                      x 2 a 2 − y 2 b 2 = 1

This equation defines a hyperbola centered at the origin with vertices   ( ± a , 0 )   and co-vertices   ( 0 ± b ) .

Standard forms of the equation of a hyperbola with center (0,0)

The standard form of the equation of a hyperbola with center   ( 0 , 0 )   and transverse axis on the x -axis is

x 2 a 2 − y 2 b 2 = 1

where

  • the length of the transverse axis is   2 a
  • the coordinates of the vertices are   ( ± a , 0 )
  • the length of the conjugate axis is   2 b
  • the coordinates of the co-vertices are   ( 0, ± b )
  • the distance between the foci is   2 c , where   c 2 = a 2 + b 2
  • the coordinates of the foci are   ( ± c , 0 )
  • the equations of the asymptotes are   y = ± b a x

See [link] a .

The standard form of the equation of a hyperbola with center   ( 0 , 0 )   and transverse axis on the y -axis is

y 2 a 2 − x 2 b 2 = 1

where

  • the length of the transverse axis is   2 a
  • the coordinates of the vertices are   ( 0, ± a )
  • the length of the conjugate axis is   2 b
  • the coordinates of the co-vertices are   ( ± b , 0 )
  • the distance between the foci is   2 c , where   c 2 = a 2 + b 2
  • the coordinates of the foci are   ( 0, ± c )
  • the equations of the asymptotes are   y = ± a b x

See [link] b .

Note that the vertices, co-vertices, and foci are related by the equation   c 2 = a 2 + b 2 .   When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci.

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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