Short questions

  • Thread starter dynamic998
  • Start date
  • #1
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?
 

Answers and Replies

  • #2
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.
 
  • #3
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
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.
 

Related Threads on Short questions

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
11
Views
3K
Replies
4
Views
1K
  • Last Post
Replies
6
Views
676
  • Last Post
Replies
2
Views
1K
Replies
7
Views
1K
Replies
3
Views
4K
Replies
2
Views
2K
Replies
5
Views
3K
Top