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Question-200268

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Question Number 200268 by cortano12 last updated on 16/Nov/23
Answered by ajfour last updated on 16/Nov/23
D origin. O_2 ≡(s−r, s−r) O_3 ≡(r, s−r) let eqn of line DF be y=mx r^2 =(((s−r)^2 (1−m)^2 )/(1+m^2 ))=(({(s−r)−mr}^2 )/(1+m^2 )) let (r/s)=t t^2 (1+m^2 )=(1−t)^2 (1−m)^2 ={(1−t)−mt}^2 ⇒ t^2 =(1−t)^2 −2mt(1−t) ...(i) & (1−t)(1−m)=−(1−t−mt) ⇒ 2−2t−m=0 ⇒ m=2(1−t) substituting in ..(i) 2t−1+4t(1−t)^2 =0 8t^3 −16t^2 +12t−2=0 2t=z z^3 −4z^2 +6z−2=0 z=((4+(3(√(33))−17)^(1/3) −(3(√(33))+17)^(1/3) )/3) ≈ 0.456311 t=(r/s)=(z/2)≈0.228155
$${D}\:{origin}. \\ $$$${O}_{\mathrm{2}} \equiv\left({s}−{r},\:{s}−{r}\right) \\ $$$${O}_{\mathrm{3}} \equiv\left({r},\:{s}−{r}\right) \\ $$$${let}\:\:{eqn}\:{of}\:{line}\:{DF}\:{be}\: \\ $$$${y}={mx} \\ $$$${r}^{\mathrm{2}} =\frac{\left({s}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}−{m}\right)^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} }=\frac{\left\{\left({s}−{r}\right)−{mr}\right\}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$${let}\:\frac{{r}}{{s}}={t} \\ $$$${t}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)=\left(\mathrm{1}−{t}\right)^{\mathrm{2}} \left(\mathrm{1}−{m}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left\{\left(\mathrm{1}−{t}\right)−{mt}\right\}^{\mathrm{2}} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} =\left(\mathrm{1}−{t}\right)^{\mathrm{2}} −\mathrm{2}{mt}\left(\mathrm{1}−{t}\right)\:\:…\left({i}\right) \\ $$$$\&\:\:\left(\mathrm{1}−{t}\right)\left(\mathrm{1}−{m}\right)=−\left(\mathrm{1}−{t}−{mt}\right) \\ $$$$\Rightarrow\:\:\mathrm{2}−\mathrm{2}{t}−{m}=\mathrm{0} \\ $$$$\Rightarrow\:\:{m}=\mathrm{2}\left(\mathrm{1}−{t}\right) \\ $$$${substituting}\:{in}\:..\left({i}\right) \\ $$$$\mathrm{2}{t}−\mathrm{1}+\mathrm{4}{t}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{8}{t}^{\mathrm{3}} −\mathrm{16}{t}^{\mathrm{2}} +\mathrm{12}{t}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{t}={z} \\ $$$${z}^{\mathrm{3}} −\mathrm{4}{z}^{\mathrm{2}} +\mathrm{6}{z}−\mathrm{2}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{4}+\left(\mathrm{3}\sqrt{\mathrm{33}}−\mathrm{17}\right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{3}\sqrt{\mathrm{33}}+\mathrm{17}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}} \\ $$$$\approx\:\mathrm{0}.\mathrm{456311} \\ $$$${t}=\frac{{r}}{{s}}=\frac{{z}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{228155} \\ $$

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