Question

If $(x-a{)}^2+(y-b{)}^2=c^2$, then prove that $\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}{\frac{d^{2}y}{d{x}^2}}$ is a independent of $C$.

Answer 1

Given,

$(x-a{)}^2+(y-b{)}^2=c^2....\left(i\right)$

Diff. w. r. t $x$

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

$\Rightarrow \frac{dy}{dx}=\frac{(a-x)}{y-b}....\left(ii\right)$

Again Diff. w. r. $x$

$\frac{d^{2}y}{d{x}^2}=\frac{(0-1)(y-b)-(\frac{dy}{dx}-0)(a-x)}{(y-b{)}^2}$ $[\because \frac{d}{dx}\frac{v}{u}=\frac{u\frac{dv}{dx}-v\frac{du}{dx}}{u^2}]$

$=\frac{(-y+b)-(a-x)\frac{dy}{dx}}{(y-b{)}^2}$

$=-\frac{(y-b)-(x-a)\left(\frac{-(x-a)}{y-b}\right)}{(y-b{)}^2}$ [From (1)]

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^3}$

$\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{\frac{3}{2}}}{\frac{d^{2}y}{d{x}^2}}$

$=\frac{{[1+{\left(\frac{(x-a)}{y-b}\right)}^2]}^{\frac{3}{2}}}{\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^3}}$

$=\frac{{\left[\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^2}\right]}^{\frac{3}{2}}}{\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^2}\times \frac{1}{(y-b)}}$

$=\frac{{\left[\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^2}\right]}^{\frac{2}{2}}}{\frac{1}{(y-b)}}$

$=\left[\frac{(y-b{)}^2+(x-a{)}^2}{(y-b{)}^2}\right]\times \frac{y-b}{1}$

$=\left[\frac{(y-b{)}^2+(x-a{)}^2}{(y-b)}\right]$

$=(y-b)+\frac{(x-a{)}^2}{y-b}$

Hence, it is independent of $C$

Hence proved.