short questions

[dynamic998]

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?

[KLscilevothma]

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.

[Integral]

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.