If x = 2bt/(1 + t^2), y = a((1 - t^2)/(1 + t^2)), show that dy/dx = -(b^2 y)/a^2 x

Question

If x = \(\frac{2bt}{1 + t^2}, \,y = a\left(\frac{1 - t^2}{1 + t^2}\right),\) show that dy/dx = \(-\frac{b^2 y}{a^2 x}\)

x = 2bt/(1 + t²), y = a((1 - t²)/(1 + t²)),

Answer 1

x = 2bt/(1 + t²), y = a((1 - t²)/(1 + t²))

Differentiating both sides w.r.t. y, we get