<< 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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 4

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




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