Question

If ${ax}^2+2xy+{by}^2=0$ then find $\frac{dy}{dx}.$

Answer 1

$a{x}^2+2xy+b{y}^2=0$

$\Rightarrow a\frac{d}{dx}\left(x^2\right)+2\frac{d}{dx}\left(xy\right)+b\frac{d}{dx}\left(y^2\right)=0$

$\Rightarrow a\left(2x\right)+2(x\frac{dy}{dx}+y)+b\left(2y\frac{dy}{dx}\right)=0$

$\Rightarrow 2ax+2x\frac{dy}{dx}+2by\frac{dy}{dx}=0$

$\Rightarrow (2x+2by)\frac{dy}{dx}=-2ax$

$\Rightarrow \frac{dy}{dx}=\frac{-2ax}{2(x+by)}$

$\therefore \frac{dy}{dx}=\frac{-ax}{x+by}$