Question

If $a{x}^2+2hxy+b{y}^2=0$, find $\frac{dy}{dx}$ & $\frac{d^{2}y}{dx}$

Answer 1

$a{x}^2+2hxy+b{y}^2=0$
$2ax+2h[x\frac{dy}{dx}+y]+2yb\frac{dy}{dx}=0$
$\frac{dy}{dx}[hx+yb]=-(hy+ax)$
$\frac{dy}{dx}=-\frac{(hy+ax)}{(hx+yb)}$
$\frac{d^{2}y}{d{x}^2}=\frac{-(hx+yb)[h\frac{dy}{dx}+a]+(hy+ax)[h+b\frac{dy}{dx}]}{{(hx+yb)}^2}$
$=\frac{(hy+ax)[h-\frac{b(hy+ax)}{hx+yb}]-(hx+yb)[-h\frac{(hy+ax)}{(hx+yb)}+a]}{{(hx+yb)}^2}$
$=\frac{(hy+ax)(h^{2}x+hyb-bhy-abx)-(hx+yb)(-h^{2}y-hax+hax+aby)}{{(hx+yb)}^3}$

$=\frac{h^{3}xy-habxy+a{h}^2{x}^2-a^{2}b{x}^2-h^{3}xy+haby-a{h}^2{x}^2+a^{2}b{x}^2}{{(hx+yb)}^3}$

$=0$