Variation

y ∝ x, then y = kx

x² + y²
= x² + (kx)²
= x² + k²x²
= (1 + k²)x²
= [(1 + k²)k]*x*(kx)
= [(1 + k²)k]xy
= constant*xy

Since x² + y² = constant*xy
then x² + y² ∝ xy

2011-06-17 01:56:55 補充：
打漏了除號。

第六行和第七行應是：
= [(1 + k²)/k]*x*(kx)
= [(1 + k²)/k]xy

y varies directly as x
=> y = kx, where k is a constant.

x^2 + y^2
= x^2 + (kx)^2
= (1 + k^2) x^2
= (1 + k^2) * x * x
= (1 + k^2) * x * (y/k)
= (1 + k^2)/k * xy (k should not be zero)
= cxy

where c = (1 + k^2)/k is also a constant.

The statement "If y varies directly as x, is (x^2 + y^2) also varies directly as xy?" is true.

2011-06-16 23:59:28 補充：
andrew, there is a mistake made in your answer.
The constant should be equal to (1+k^2)/k, not (1+k^2)k