Question-70738

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Question Number 70738 by ajfour last updated on 07/Oct/19

Commented by ajfour last updated on 07/Oct/19

$${Find}\:\:\frac{{R}}{{a}}\:\centerdot \\ $$

Commented by ajfour last updated on 07/Oct/19

Answered by mr W last updated on 08/Oct/19

C(k,R) P(−h,h^2 ) y′=2x=−2h eqn. of PC: y=((x+h)/(2h))+h^2 R=((k+h)/(2h))+h^2 ⇒k+h=2h(R−h^2 ) (k+h)^2 +(R−h^2 )^2 =R^2 4h^2 (R−h^2 )^2 +(R−h^2 )^2 =R^2 4R^2 −8Rh^2 +4h^4 −2R+h^2 =0 4h^4 −(8R−1)h^2 +2R(2R−1)=0 h^2 =((8R−1−(√(16R+1)))/8) ⇒h=(1/2)(√((8R−1−(√(16R+1)))/2)) ⇒k=[2(R−h^2 )−1]h center of small circle A(−r,p) (k+r)^2 +(R−p)^2 =(R+r)^2 (R−p)^2 =(R−k)(R+k+2r) p=R−(√((R−k)(R+k+2r))) T(−s,s^2 ) y′=−2s eqn. of TA: y=((x+s)/(2s))+s^2 p=((−r+s)/(2s))+s^2 p−s^2 =((s−r)/(2s)) (−r+s)^2 +(p−s^2 )^2 =r^2 4s^4 −8rs^3 +s^2 −2rs+r^2 =0 r^2 −2s(4s^2 +1)r+s^2 (4s^2 +1)=0 ⇒r=s(4s^2 +1)−2s^2 (√(4s^2 +1)) ...(i) p=s^2 +((s−r)/(2s))=R−(√((R−k)(R+k+2r))) ...(ii) we get s from (ii) for given R, and then r from (i). examples: R=8 ⇒r=0.6351 R=4 ⇒r=0.2956

$${C}\left({k},{R}\right) \\ $$$${P}\left(−{h},{h}^{\mathrm{2}} \right) \\ $$$${y}'=\mathrm{2}{x}=−\mathrm{2}{h} \\ $$$${eqn}.\:{of}\:{PC}: \\ $$$${y}=\frac{{x}+{h}}{\mathrm{2}{h}}+{h}^{\mathrm{2}} \\ $$$${R}=\frac{{k}+{h}}{\mathrm{2}{h}}+{h}^{\mathrm{2}} \\ $$$$\Rightarrow{k}+{h}=\mathrm{2}{h}\left({R}−{h}^{\mathrm{2}} \right) \\ $$$$\left({k}+{h}\right)^{\mathrm{2}} +\left({R}−{h}^{\mathrm{2}} \right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{4}{h}^{\mathrm{2}} \left({R}−{h}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({R}−{h}^{\mathrm{2}} \right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} −\mathrm{8}{Rh}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{4}} −\mathrm{2}{R}+{h}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}{h}^{\mathrm{4}} −\left(\mathrm{8}{R}−\mathrm{1}\right){h}^{\mathrm{2}} +\mathrm{2}{R}\left(\mathrm{2}{R}−\mathrm{1}\right)=\mathrm{0} \\ $$$${h}^{\mathrm{2}} =\frac{\mathrm{8}{R}−\mathrm{1}−\sqrt{\mathrm{16}{R}+\mathrm{1}}}{\mathrm{8}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{8}{R}−\mathrm{1}−\sqrt{\mathrm{16}{R}+\mathrm{1}}}{\mathrm{2}}} \\ $$$$\Rightarrow{k}=\left[\mathrm{2}\left({R}−{h}^{\mathrm{2}} \right)−\mathrm{1}\right]{h} \\ $$$$ \\ $$$${center}\:{of}\:{small}\:{circle}\:{A}\left(−{r},{p}\right) \\ $$$$\left({k}+{r}\right)^{\mathrm{2}} +\left({R}−{p}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} \\ $$$$\left({R}−{p}\right)^{\mathrm{2}} =\left({R}−{k}\right)\left({R}+{k}+\mathrm{2}{r}\right) \\ $$$${p}={R}−\sqrt{\left({R}−{k}\right)\left({R}+{k}+\mathrm{2}{r}\right)} \\ $$$$ \\ $$$${T}\left(−{s},{s}^{\mathrm{2}} \right) \\ $$$${y}'=−\mathrm{2}{s} \\ $$$${eqn}.\:{of}\:{TA}: \\ $$$${y}=\frac{{x}+{s}}{\mathrm{2}{s}}+{s}^{\mathrm{2}} \\ $$$${p}=\frac{−{r}+{s}}{\mathrm{2}{s}}+{s}^{\mathrm{2}} \\ $$$${p}−{s}^{\mathrm{2}} =\frac{{s}−{r}}{\mathrm{2}{s}} \\ $$$$\left(−{r}+{s}\right)^{\mathrm{2}} +\left({p}−{s}^{\mathrm{2}} \right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{4}{s}^{\mathrm{4}} −\mathrm{8}{rs}^{\mathrm{3}} +{s}^{\mathrm{2}} −\mathrm{2}{rs}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{s}\left(\mathrm{4}{s}^{\mathrm{2}} +\mathrm{1}\right){r}+{s}^{\mathrm{2}} \left(\mathrm{4}{s}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}={s}\left(\mathrm{4}{s}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}{s}^{\mathrm{2}} \sqrt{\mathrm{4}{s}^{\mathrm{2}} +\mathrm{1}}\:\:\:…\left({i}\right) \\ $$$${p}={s}^{\mathrm{2}} +\frac{{s}−{r}}{\mathrm{2}{s}}={R}−\sqrt{\left({R}−{k}\right)\left({R}+{k}+\mathrm{2}{r}\right)}\:\:\:…\left({ii}\right) \\ $$$${we}\:{get}\:{s}\:{from}\:\left({ii}\right)\:{for}\:{given}\:{R}, \\ $$$${and}\:{then}\:{r}\:{from}\:\left({i}\right). \\ $$$${examples}: \\ $$$${R}=\mathrm{8}\:\Rightarrow{r}=\mathrm{0}.\mathrm{6351} \\ $$$${R}=\mathrm{4}\:\Rightarrow{r}=\mathrm{0}.\mathrm{2956} \\ $$

Commented by mr W last updated on 08/Oct/19