Question

If $\sqrt{1-x^6}+\sqrt{1-y^6}=a(x^3-y^3)$ and $\frac{dy}{dx}=f(x,y)\sqrt{\frac{1-y^6}{1-x^6}}$ then

• A. $f(x,y)=\frac{y}{x}$
• B. $f(x,y)=\frac{x^2}{y^2}$
• C. $f(x,y)=2\frac{y^2}{x^2}$
• D. $f(x,y)=\frac{y^2}{x^2}$

Answer 1

Answer: (B)

$f(x,y)=\frac{x^2}{y^2}$

$\sqrt{1-x^6}+\sqrt{1-y^6}=a(x^3-y^3)$

$\Rightarrow -\frac{6\times x^5}{2\sqrt{{1-x}^6}}-\frac{6\times y^5}{2\sqrt{1-y^6}}\frac{dy}{dx}=3a(x^2-y^2\frac{dy}{dx})$

$\Rightarrow (-\frac{3y^5}{\sqrt{1-y^6}}+{3ay}^2)\frac{dy}{dx}=3a{x}^2+\frac{{3x}^5}{\sqrt{{1-x}^6}}$
$\Rightarrow \frac{dy}{dx}=\frac{a{x}^2+\frac{x^5}{\sqrt{{1-x}^6}}}{{ay}^2-\frac{y^5}{\sqrt{{1-y}^6}}}$

$\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\left(\frac{a+\frac{x^3}{\sqrt{{1-x}^6}}}{a-\frac{y^3}{\sqrt{{1-y}^6}}}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\left(\frac{\frac{\sqrt{1-x^6}+\sqrt{1-y^6}}{x^3-y^3}+\frac{x^3}{\sqrt{{1-x}^6}}}{\frac{\sqrt{1-x^6}+\sqrt{1-y^6}}{x^3-y^3}-\frac{y^3}{\sqrt{{1-y}^6}}}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\left(\frac{\frac{1-x^6+\sqrt{(1-y^6)(1-x^6)}+x^6-x^3{y}^3}{\sqrt{{1-x}^6}}}{\frac{\sqrt{(1-x^6)(1-y^6)}+1-y^6-x^3{y}^3+y^6}{\sqrt{{1-y}^6}}}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\sqrt{\frac{{1-y}^6}{{1-x}^6}}$

Hence, $f(x,y)=\frac{x^2}{y^2}$