Prove the following

Question:

Let $x^{k}+y^{k}=a^{k},(a, K>0)$ and $\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0$,

then $\mathrm{k}$ is :

  1. $\frac{3}{2}$

  2. $\frac{1}{3}$

  3. $\frac{2}{3}$

  4. $\frac{4}{3}$


Correct Option: , 3

Solution:

$x^{k}+y^{k}=a^{k}(a, k>0)$

$k x^{k-1}+k y^{k-1} \frac{d y}{d x}=0$

$\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{k}-1}=0 \Rightarrow \mathrm{k}-1=-\frac{1}{3} \Rightarrow \mathrm{k}=2 / 3$

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