X^2-Y^2=k(X^2+Y^2)

X2 - Y2 = k(X+ Y)2
(X+Y)(X-Y) = k(X+ Y)2

Suppose X is not equal to -Y (so X+Y is non-zero),
we can divide (X+Y) from both side to get
(X-Y) = k(X+Y)

2007-04-22 18:06:54 補充：
我和閣下意見一樣認為題目有錯我認為原題目應該是(X+Y)^2

2007-04-23 09:45:12 補充：
正確答案X^2-Y^2=k(X^2+Y^2)(1-k)X^2=(1+k)Y^2Y^2=(1-k)/(1+k)X^2Y= [(1-k)/(1+k)]^(1/2)X or Y= -[(1-k)/(1+k)]^(1/2)Xi.e. Y is directly proportional to X

2007-04-23 09:47:46 補充：
part BSince Y is directly proportional to XY = k'X for a constant k'X - Y = (1-k')XX + Y = (1+k')Xwe can express X - Y as [ (1-k')/(1+k') ](X + Y)let k'' = (1-k')/(1+k')X - Y = k'(X + Y)i.e. X - Y is directly proportional to X + Y

樓主呢題係邊本練習書揾架，好似出錯題：

X^2-Y^2=k(X^2+Y^2)

(1-k)X^2=(1+k)Y^2

Y^2=(1-k)/(1+k)X^2

Y= X((1-k)/(1+k))^0.5 or Y= - X((1-k)/(1+k))^0.5

題目要証y=kx

即係要証
+
k= ((1-k)/(1+k))^0.5
-
咁即係要証:
k^2=1-k^2

k^2=1/2

k=+
(1/2)^0.5
_
所以呢題數要有一些特定value既k,
y=kx先會成立
但係因為題目冇givenｋ既值
所以係証唔到

2007-04-22 11:16:50 補充：
上便嗰位人兄做乜改咗人地條題目ｘ＾２＋ｙ＾２冇端端點會等於（ｘ＋ｙ）＾２