<< Chapter < Page Chapter >> Page >

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.

Questions & Answers

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
Kc Reply
1+cos²A/cos²A=2cosec²A-1
Ramesh Reply
test for convergence the series 1+x/2+2!/9x3
success Reply
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
Lhorren Reply
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
jancy Reply
answer
Ajith
exponential series
Naveen
what is subgroup
Purshotam Reply
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
Macmillan Reply
e power cos hyperbolic (x+iy)
Vinay Reply
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
Payal Reply
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
Tejas Reply
why {2kπ} union {kπ}={kπ}?
Huy Reply
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
Trilochan Reply
what is complex numbers
Ayushi Reply
Please you teach
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Anshuman Reply
Solve 2cos x + 3sin x = 0.5
shobana Reply
Practice Key Terms 4

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

Ask