Question

Form the differential equation by eliminating the arbitrary constant $a$ from the relation ${(x-a)}^2+y^2=1$.

Answer 1

$(x-a{)}^2+y^2$ = 1

$x-a=\sqrt{1-y^2}$

$a=x-\sqrt{1-y^2}$

Diffrntiating with respect to x

$0=1-\frac{1}{2\sqrt{1-y^2}}(-2y)\frac{dy}{dx}$ ...$[\frac{d\sqrt{x}}{dx}=\frac{1}{2\sqrt{2}}\frac{d}{dx}{x}^n=n{x}^{n-1}\frac{d}{dx}(y)=\frac{dy}{dx}]$

$\Rightarrow 0=1+\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx}$

$\Rightarrow -1=\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{y}$