If $x^k+y^k=a^k(a, k0)$ and $\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}-0$, then $\mathrm{k}$ has the value

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If $x^k+y^k=a^k(a, k>0)$ and $\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}-0$, then $\mathrm{k}$ has the value

Aicharya · Accepted Answer

$x^{\mathrm{k}}+y^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}}$ Differentiating w.r.t. $x$, we get $\begin{aligned} & \mathrm{k} x^{k-1}+\mathrm{k} y^{\mathrm{k}-1} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \ 	herefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\mathrm{k} x^{\mathrm{k}-1}}{\mathrm{k} y^{\mathrm{k}-1}} \ 	herefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\left(\frac{x}{y}ight)^{\mathrm{k}-1} \ & \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}ight)^{1-\mathrm{k}}=0 \ & 	ext { But } \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}ight)^{\frac{1}{3}}=0 ...[Given] \end{aligned}$ Comparing above equations, we get $\begin{aligned} 1-k & =\frac{1}{3} \ & 1-\frac{1}{3}=k \ 	herefore \quad k & =\frac{2}{3} \end{aligned}$

If $x^k+y^k=a^k(a, k>0)$ and $\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}-0$, then $\mathrm{k}$…

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