Question

If (x-a)2+(y-b)2=c2, for some c>0 prove that
[1+($\frac{dy}{dx}$)2]$^{\frac{3}{2}}$/$\frac{d^2y}{dx^2}$ is a constant independent of a and b

Answer 1

It is given that,(x-a)2+(y-b)2=c2
Differentiating both sides with respect to x, we obtain
$\frac{d}{dx}$[(x-a)2]+$\frac{d}{dx}$[(y-b)2]=$\frac{d}{dx}$(c2)
⇒ 2(x-a).$\frac{d}{dx}$(x-a)+2(y-b).$\frac{d}{dx}$(y-b)=0
⇒2(x-a).1+2(y-b).$\frac{dy}{dx}$=0
⇒ $\frac{dy}{dx}$=$\frac{-(x-a)}{y-b}$ ...(1)
∴$\frac{d^2y}{dx^2}$=$\frac{d}{dx}$[$\frac{-(x-a)}{y-b}$]
=-[(y-b).$\frac{d}{dx}$(x-a)-(x-a).\frac{d}{dx}$$\frac{(y-b)}{(y-b)^2}]
=-c, where c is constant and is independent of a and b
Hence, proved.