Question

If $\sqrt{1-x^4}+\sqrt{1-y^4}=k(x^2-y^2),$ prove that $\frac{dy}{dx}=\frac{x\sqrt{1-y^4}}{y\sqrt{1-x^4}}$

Answer 1

Given,

$\sqrt{1-x^4}+\sqrt{1-y^4}=k(x^2-y^2)$

Differentiate on both sides w.r.t k

$\frac{-4x^3}{2\sqrt{1-x^4}}-\frac{4y^3}{2\sqrt{1-y^4}}\left(\frac{dy}{dx}\right)=k(2x-2y\frac{dy}{dx})$

$2kx+\frac{4x^3}{2\sqrt{1-x^4}}=(2ky-\frac{4y^3}{2\sqrt{1-y^4}})\frac{dy}{dx}$

$kx+\frac{x^3}{\sqrt{1-x^4}}=(ky-\frac{y^3}{\sqrt{1-y^4}})\frac{dy}{dx}$

$\frac{x\sqrt{1-x^4}+x\sqrt{1-x^4}}{x^2-y^2}+\frac{x^3}{\sqrt{1-x^4}}=(\frac{y\sqrt{1-y^4}+y\sqrt{1-y^4}}{x^2-y^2}+\frac{y^3}{\sqrt{1-y^4}})\frac{dy}{dx}$

$\frac{x-x^5+x\sqrt{(1-y^4)(1-x^4)}+x^5-x^3{y}^2}{\sqrt{1-x^4}}=\left(\frac{y\sqrt{(1-y^4)(1-x^4)}+y-y^5-x^2{y}^3+y^5}{\sqrt{1-y^4}}\right)\frac{dy}{dx}$

$\frac{x(1-\sqrt{(1-y^4)(1-x^4)}-x^2{y}^2)}{\sqrt{1-x^4}}=\frac{y(\sqrt{(1-y^4)(1-x^4)}+1-x^2{y}^2)}{\sqrt{1-y^4}}\frac{dy}{dx}$

$\therefore \frac{dy}{dx}=\frac{x\sqrt{1-y^4}}{y\sqrt{1-x^4}}$