1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Short questions

  1. Apr 13, 2003 #1
    x/2 = 3/y. Can anyone explain why this is an hyperbola?

    find the third term of the expansion (2x-y)to the third power.
    I know u have to do the things with the combinations but all i get is 20xy² but the answer is 6xy². Can anyone help?
  2. jcsd
  3. Apr 13, 2003 #2
    I think you mean to find the 3rd term in descending powers of x

    Method 1:
    (2x-y)3 = (Summation r from 0 to 3) C3r (2x)r (-y)3-r

    So the third term

    Method 2:
    You can expand (2x-y)3 directly.
  4. Apr 13, 2003 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    To see that your first problem

    x/2 = 3/y is a hyperbola, do a coordinate transform to rotate the coordinate axis by π/2 radians.

    you will find that (let u & v be the new axis)

    u=xcosθ + ysinθ
    v=-xsinθ + ycosθ

    Let θ = π/2

    solve for x & y

    x = (v-u)/sqrt(2) y=(v+u)/sqrt(2)

    substituting this back into the origianal relationship gives

    (v-u)(v+u)/2 =6


    v*v - u*u = 12

    This is the standard form for a hyperbola.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook