Question

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

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

Answer 1

The correct option is D

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

Explanation for the correct option:

Step 1. Put $x^3=\cos \theta$ and $y^3=\cos ɸ$ in given equation, we get

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

$\Rightarrow$$\sqrt{1–{\cos }^2\theta }+\sqrt{1–{\cos }^2ɸ}=a(\cos \theta –\cos ɸ)$

$\Rightarrow$ $\sin \theta +\sin ɸ=a(\cos \theta –\cos ɸ)$

$\Rightarrow$ $2sin\frac{[\theta +ɸ]}{2}cos\frac{[\theta –ɸ]}{2}=–2asin\frac{[\theta +ɸ]}{2}sin\frac{[\theta –ɸ]}{2}$

$\Rightarrow$ $cot\frac{[\theta –ɸ]}{2}=–a$

$\Rightarrow$ $[\theta –ɸ]=2co{t}^{-1}(-a)$ (Constant)

Step 2. Differentiate the given equation,

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

$–\frac{3x^2}{\sqrt{1–x^6}}+\frac{3y^2}{\sqrt{1–y^6}}\frac{dy}{dx}=0$

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

Comparing this with $\frac{dy}{dx}=f(x,y)\sqrt{\left(\frac{1-y^6}{1-x^6}\right)}$

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

Hence , Option ‘D’ is Correct.