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Question Number 24555 by ajfour last updated on 20/Nov/17

Commented by ajfour last updated on 20/Nov/17

$$Findthemaximumradiusof$$$$acirclesuchthatittouchesthe$$$$parabolay={Ax}^{2}atitsvertex,$$$$andliesaboveitandcutsit$$$$nowhere.$$

Commented by mrW1 last updated on 21/Nov/17

$$y={Ax}^2\left(parabola\right)$$$$x^2+{(y-r)}^2=r^2\left(circle\right)$$$$y+A{(y-r)}^2-{Ar}^2=0$$$$\Rightarrow A{(y-r)}^2+(y-r)-r(Ar-1)=0$$$$D=1+4Ar(Ar-1)=0$$$${(2Ar-1)}^2=0$$$$\Rightarrow r=\frac{1}{2A}$$

Commented by ajfour last updated on 20/Nov/17

$$thankyousir.$$

Answered by ajfour last updated on 20/Nov/17

$$x^2+{(y-r)}^2=r^2$$$$2x+2(y-r)\frac{dy}{dx}=0$$$$1+{\left(\frac{dy}{dx}\right)}^2+(y-r)\frac{d^{2}y}{{dx}^2}=0$$$$atx=0,\frac{dy}{dx}=0,So$$$$\frac{d^{2}y}{{dx}^2}=\frac{1}{r-y}=\frac{1}{r}=\frac{d^2}{{dx}^2}\left({Ax}^2\right)$$$$\Rightarrow r=\frac{1}{2A}.$$

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