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

Commented by ajfour last updated on 20/Nov/17

Find the maximum radius of  a circle such that it touches the  parabola y=Ax^2  at its vertex,  and lies above it and cuts it  nowhere.

$${Find}\:{the}\:{maximum}\:{radius}\:{of} \\ $$$${a}\:{circle}\:{such}\:{that}\:{it}\:{touches}\:{the} \\ $$$${parabola}\:{y}={Ax}^{\mathrm{2}} \:{at}\:{its}\:{vertex}, \\ $$$${and}\:{lies}\:{above}\:{it}\:{and}\:{cuts}\:{it} \\ $$$${nowhere}. \\ $$

Commented by mrW1 last updated on 21/Nov/17

y=Ax^2      (parabola)  x^2 +(y−r)^2 =r^2     (circle)  y+A(y−r)^2 −Ar^2 =0  ⇒A(y−r)^2 +(y−r)−r(Ar−1)=0  D=1+4Ar(Ar−1)=0  (2Ar−1)^2 =0  ⇒r=(1/(2A))

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

Commented by ajfour last updated on 20/Nov/17

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Answered by ajfour last updated on 20/Nov/17

x^2 +(y−r)^2 =r^2   2x+2(y−r)(dy/dx)=0  1+((dy/dx))^2 +(y−r)(d^2 y/dx^2 ) =0  at x=0 , (dy/dx)=0 , So  (d^2 y/dx^2 ) =(1/(r−y)) =(1/r) = (d^2 /dx^2 )(Ax^2 )  ⇒  r=(1/(2A)) .

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

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