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.