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Short questionsby dynamic998
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#1
Apr1303, 01:21 AM

P: n/a

x/2 = 3/y. Can anyone explain why this is an hyperbola?
find the third term of the expansion (2xy)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? 


#2
Apr1303, 04:07 AM

P: 321

Method 1: (2xy)^{3} = (Summation r from 0 to 3) C^{3}_{r} (2x)^{r} (y)^{3r} So the third term =C^{3}_{2}(2x)(y)^{2} =6xy^{2} Method 2: You can expand (2xy)^{3} directly. 


#3
Apr1303, 05:53 PM

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P: 7,334

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 = (vu)/sqrt(2) y=(v+u)/sqrt(2) substituting this back into the origianal relationship gives (vu)(v+u)/2 =6 or v*v  u*u = 12 This is the standard form for a hyperbola. 


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