Question-182298

Question Number 182298 by SEKRET last updated on 07/Dec/22

Answered by mr W last updated on 07/Dec/22

radius: r center: (k,2−r) touching point with y=x^2 : (p,p^2 ) touching point with y=(x^2 /4): (q,(q^2 /4)) tan θ=2p k=p+r sin θ ⇒k=p+((2pr)/( (√(1+4p^2 )))) 2−r=p^2 −r cos θ ⇒2−r=p^2 −(r/( (√(1+4p^2 )))) tan ϕ=(q/2) k=q−r sin ϕ ⇒k=q−((qr)/( (√(4+q^2 )))) 2−r=(q^2 /4)+r cos ϕ ⇒2−r=(q^2 /4)+((2r)/( (√(4+q^2 )))) ⇒r=((q−p)/(((2p)/( (√(1+4p^2 ))))+(q/( (√(4+q^2 )))))) ⇒r=((2−p^2 )/(1−(1/( (√(1+4p^2 )))))) ⇒r=((2−(q^2 /4))/(1+(2/( (√(4+q^2 )))))) ⇒p≈1.29415, q≈2.13965 ⇒r≈0.5084

$${radius}:\:{r} \\ $$$${center}:\:\left({k},\mathrm{2}−{r}\right) \\ $$$${touching}\:{point}\:{with}\:{y}={x}^{\mathrm{2}} :\:\left({p},{p}^{\mathrm{2}} \right) \\ $$$${touching}\:{point}\:{with}\:{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}:\:\left({q},\frac{{q}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$${k}={p}+{r}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{k}={p}+\frac{\mathrm{2}{pr}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\mathrm{2}−{r}={p}^{\mathrm{2}} −{r}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{2}−{r}={p}^{\mathrm{2}} −\frac{{r}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\varphi=\frac{{q}}{\mathrm{2}} \\ $$$${k}={q}−{r}\:\mathrm{sin}\:\varphi \\ $$$$\Rightarrow{k}={q}−\frac{{qr}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }} \\ $$$$\mathrm{2}−{r}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+{r}\:\mathrm{cos}\:\varphi \\ $$$$\Rightarrow\mathrm{2}−{r}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow{r}=\frac{{q}−{p}}{\frac{\mathrm{2}{p}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}+\frac{{q}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}−{p}^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}−\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}{\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}+{q}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{29415},\:{q}\approx\mathrm{2}.\mathrm{13965} \\ $$$$\Rightarrow{r}\approx\mathrm{0}.\mathrm{5084} \\ $$

Commented by mr W last updated on 07/Dec/22