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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.

Questions & Answers

for the "hiking" mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. if there is the same amount of almonds as cashews, how many of each item is in the trail mix?
ADNAN Reply
linear speed of an object
Melissa Reply
an object is traveling around a circle with a radius of 13 meters .if in 20 seconds a central angle of 1/7 Radian is swept out what are the linear and angular speed of the object
Melissa
test
Matrix
how to find domain
Mohamed Reply
like this: (2)/(2-x) the aim is to see what will not be compatible with this rational expression. If x= 0 then the fraction is undefined since we cannot divide by zero. Therefore, the domain consist of all real numbers except 2.
Dan
define the term of domain
Moha
if a>0 then the graph is concave
Angel Reply
if a<0 then the graph is concave blank
Angel
what's a domain
Kamogelo Reply
The set of all values you can use as input into a function su h that the output each time will be defined, meaningful and real.
Spiro
how fast can i understand functions without much difficulty
Joe Reply
what is inequalities
Nathaniel
functions can be understood without a lot of difficulty. Observe the following: f(2) 2x - x 2(2)-2= 2 now observe this: (2,f(2)) ( 2, -2) 2(-x)+2 = -2 -4+2=-2
Dan
what is set?
Kelvin Reply
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
Divya Reply
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
Thomas
158.5 This number can be developed by using algebra and logarithms. Begin by moving log(2) to the right hand side of the equation like this: t/100 log(2)= log(3) step 1: divide each side by log(2) t/100=1.58496250072 step 2: multiply each side by 100 to isolate t. t=158.49
Dan
what is the importance knowing the graph of circular functions?
Arabella Reply
can get some help basic precalculus
ismail Reply
What do you need help with?
Andrew
how to convert general to standard form with not perfect trinomial
Camalia Reply
can get some help inverse function
ismail
Rectangle coordinate
Asma Reply
how to find for x
Jhon Reply
it depends on the equation
Robert
yeah, it does. why do we attempt to gain all of them one side or the other?
Melissa
how to find x: 12x = 144 notice how 12 is being multiplied by x. Therefore division is needed to isolate x and whatever we do to one side of the equation we must do to the other. That develops this: x= 144/12 divide 144 by 12 to get x. addition: 12+x= 14 subtract 12 by each side. x =2
Dan
whats a domain
mike Reply
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
Spiro; thanks for putting it out there like that, 😁
Melissa
foci (–7,–17) and (–7,17), the absolute value of the differenceof the distances of any point from the foci is 24.
Churlene Reply
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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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