Question

If $(x-a{)}^2+(y-b{)}^2=c^2$ prove that $\frac{{\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^{\frac{3}{2}}}{\frac{d^{2}y}{d{x}^2}}=constant$

Answer 1

We have,

${(x-a)}^2+{(y-b)}^2=c^2$ …… (1)

Differentiation this equation with respect to $x$ and we get,

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

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

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

Again differentiation and we get,

$\frac{d^{2}y}{d{x}^2}=-\frac{[(y-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(y-b)]}{{(y-b)}^2}$

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

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

Put the value of $\frac{dy}{dx}$ and we get,

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

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

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

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

Put the value of $\frac{dy}{dx}$ and $$\frac{d^{2}y}{d{x}^2}$$ in LHS and we get,

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

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

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

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

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

$=-{[{(y-b)}^2+{(x-a)}^2]}^2$ by equation (1)

$=-{\left(c^2\right)}^2=-c^4=constant$

Hence proved.