From the given equation, sinø ≠ 0, cosø ≠ 0
x = a{1 - (1/sinø)} = a(sinø - 1) / sinø ··· Eq.1
y = a{(1/cosø) + tanø} = a{(1/cosø) + (sinø/cosø)} = a(1 + sinø) / cosø, so
y² = a²(1 + sinø)² / cos²ø = a²(1 + sinø)² / (1 - sinø²) = a²(1 + sinø)² / (1+ sinø)·(1 - sinø), thus
y² = a²(1 + sinø) / (1 - sinø) ··· Eq.2
Let sinø = X. Plug X into Eq.1 and Eq.2, we have: x = a(X - 1) / X, y² = a²(1 + X) / (1 - X), so
xy² = {a(X - 1) / X}·{a²(1 + X) / (1 - X)} = -a³(1 + X) / X ··· Eq.3
a²(2a - x) = a³{2 - (X - 1) / X} = a³(X + 1) / X··· Eq.4
From Eq.3 and Eq.4, we have: xy² + a²(2a - x) = -a³{(1 + X) / X} + a³{(X + 1) / X} = 0
Therefore, xy² + a²(2a - x) = 0