Question

If $x^m+y^m=1$ where $m$ is a constant such that $\frac{dy}{dx}=-\frac{x}{y}\forall x,y\in R-\left\{0\right\},$ then the value of $m=$

Answer 1

Given:
$\frac{dy}{dx}=-\frac{x}{y}\cdots \left(1\right)$
and, $x^m+y^m=1$
Differentiating both sides with respect to $x$ we get,
$m{x}^{m-1}+m{y}^{m-1}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-{\left(\frac{x}{y}\right)}^{m-1}$
$\Rightarrow -{\left(\frac{x}{y}\right)}^{m-1}=-\frac{x}{y}\left[\text{From}\right(1\left)\right]$
$\Rightarrow m-1=1$
$\therefore m=2$