# short questions

by dynamic998
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 P: n/a 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 20xy˛ but the answer is 6xy˛. Can anyone help?
P: 321
 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 20xy?but the answer is 6xy? Can anyone help?
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.
 Mentor P: 7,208 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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