Question

If $x^2+y^2=1-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$. Show that $\frac{dy}{dx}=\frac{1}{x^{3}y}$.

Answer 1

$\begin{array}{c}{x}^2+y^2=t-\frac{1}{t}---\left(1\right)\\ x^4+y^4=t^2+\frac{1}{t^2}---\left(2\right)\\ x^4+y^4={\left(t-\frac{1}{t}\right)}^2+2\\ From\left(1\right)\\ x^4+y^4={\left(x^2+y^2\right)}^2+2\\ x^4+y^4=x^4+y^4+2x^2{y}^2+2\left[canclecommonterm\right]\\ 2x^2{y}^2=-2\\ x^2{y}^2=-1\left(3\right)\\ Differentiatingw.r.tx\\ 2x{y}^2+2y{x}^2\frac{dy}{dx}=0\\ or,\frac{dy}{dx}=\frac{-y}{x}\\ From\left(3\right)\\ \frac{dy}{dx}=\frac{y}{x}\times \frac{1}{x^2{y}^2}=\frac{1}{x^{3}y}\\ proved.\end{array}$