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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
    =C32(2x)(-y)2
    =6xy2

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

    Integral

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    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

    or

    v*v - u*u = 12

    This is the standard form for a hyperbola.
     
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