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Question Number 214483 by MATHEMATICSAM last updated on 09/Dec/24

Answered by mr W last updated on 10/Dec/24

$$circlecenterat(h,2-r)$$$$touchingy=x^{2}at(p,q)$$$$q=p^2$$$$\frac{dy}{dx}=2p=\tan \alpha$$$$h=p+r\sin \alpha =p+\frac{2pr}{\sqrt{1+4p^2}}$$$$2-r=p^2-r\cos \alpha =p^2-\frac{r}{\sqrt{1+4p^2}}$$$$2-p^2=r(1-\frac{1}{\sqrt{1+4p^2}})$$$$\{\begin{array}{l}r=\frac{2-p^2}{1-\frac{1}{\sqrt{1+4p^2}}}\\ h=p+\frac{2p(2-p^2)}{\sqrt{1+4p^2}-1}\end{array}...\left(I\right)$$$$touchingy=\frac{x^2}{4}at(u,v)$$$$v=\frac{u^2}{4}$$$$\frac{dy}{dx}=\frac{u}{2}=\tan \beta$$$$h=u-r\sin \beta =u-\frac{ur}{2\sqrt{1+\frac{u^2}{4}}}$$$$2-r=\frac{u^2}{4}+r\cos \beta =\frac{u^2}{4}+\frac{r}{\sqrt{1+\frac{u^2}{4}}}$$$$\{\begin{array}{l}r=\frac{2-\frac{u^2}{4}}{1+\frac{1}{\sqrt{1+\frac{u^2}{4}}}}\\ h=u-\frac{u(2-\frac{u^2}{4})}{2(\sqrt{1+\frac{u^2}{4}}+1)}\end{array}...\left(II\right)$$$$intersectionofbothparameterised$$$$curvesisat$$$$h\approx 1.76833$$$$r\approx 0.50832✓$$

Commented by TonyCWX08 last updated on 10/Dec/24

$$MayIknowwhythereisananglethere?$$$$I{don}^{\prime }tseethe\alpha and\beta ...$$

Commented by mr W last updated on 10/Dec/24

Commented by mr W last updated on 10/Dec/24

Commented by TonyCWX08 last updated on 11/Dec/24

$$WherecanIactuallylearnallthesestuffs??$$

Commented by ajfour last updated on 11/Dec/24

You cant find heaven anywhere other than where you are.��

Commented by mr W last updated on 11/Dec/24

$$Tolearnthingsisonething,to$$$$applythethingslearntisanother$$$$thing!$$$$Perhapsyouhavealreadylearnt$$$$allthestuffconcernedinthis$$$$question,youonlyneedtoapply$$$$it.$$

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