Question

Let $x^k+y^k=a^k$, $\left(a,k>0\right)$ and $\left(\frac{dy}{dx}\right)+{\left(\frac{y}{x}\right)}^{\frac{1}{3}}=0$, then $k$ is

• A. $\frac{1}{3}$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is C

$\frac{2}{3}$

Explanation for the correct option:

Determine the value of $k$

Let $\left(\frac{dy}{dx}\right)+{\left(\frac{y}{x}\right)}^{\frac{1}{3}}=0\to \left(i\right)$

Given, $x^k+y^k=a^k$

Differentiate the equation with respect to $x$, we get

$$\begin{gathered} \Rightarrow k{x}^{k-1}+k{y}^{k-1}\left(\frac{dy}{dx}\right)=0 \\ \Rightarrow k{x}^{k-1}=-k{y}^{k-1}\left(\frac{dy}{dx}\right) \\ \Rightarrow \left(\frac{dy}{dx}\right)=\frac{-x^{k-1}}{y^{k-1}} \\ \Rightarrow \left(\frac{dy}{dx}\right)=-{\left(\frac{y}{x}\right)}^{1-k}\left[\because \frac{x}{y}={\left(\frac{y}{x}\right)}^{-1}\right] \\ \Rightarrow \left(\frac{dy}{dx}\right)+{\left(\frac{y}{x}\right)}^{1-k}=0\to \left(ii\right) \end{gathered}$$

Equating equation $\left(i\right)\text{and}\left(ii\right)$

$$\begin{gathered} \Rightarrow \left(\frac{dy}{dx}\right)+{\left(\frac{y}{x}\right)}^{\frac{1}{3}}=\left(\frac{dy}{dx}\right)+{\left(\frac{y}{x}\right)}^{1-k} \\ \Rightarrow {\left(\frac{y}{x}\right)}^{\frac{1}{3}}={\left(\frac{y}{x}\right)}^{1-k} \\ \Rightarrow 1-k=\frac{1}{3} \\ \Rightarrow k=\frac{2}{3} \end{gathered}$$

Hence, option (C) is the correct answer