Question

If $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, then show that $\frac{dy}{dx}.\frac{dx}{dy}=1$

Answer 1

We have, $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
Differentiating write to $x,$ we get
$2ax+2h(x\frac{dy}{dx}+y)+b.2y\frac{dy}{dx}+2g+2f\frac{dy}{dx}+0=0$
Differentiating write to $y,$ we get
$2ax\frac{dx}{dy}+2h(x+y\frac{dx}{dy})+b.2y+2g\frac{dx}{dy}+2f+0=0$
Simplifying both the equations, we get
$\frac{dy}{dx}=-\frac{ax+hy+g}{hx+by+f}\frac{dx}{dy}=-\frac{hx+by+f}{ax+hy+g}\therefore \frac{dy}{dx}\times \frac{dx}{dy}=1$