Question Number 149940 by ajfour last updated on 08/Aug/21
Commented by ajfour last updated on 08/Aug/21
$${see}\:{Q}.\mathrm{149894} \\ $$
Answered by ajfour last updated on 09/Aug/21
![A(−2h+2scos θ, −2k) ; B(−2h, −2k+2ssin θ) side of eql.△ = 2s Eq. of circle_(−) x^2 +y^2 =r^2 ....(1) Eq. of DP_(−) y−ssin θ+k=cot θ(x−scos θ+h) using this in ..(1) x^2 +[hcot θ−k+ssin θ +cot θ(x−scos θ)]^2 =r^2 further x−scos θ+h=s(√3)sin θ ⇒ (r^2 /s^2 )=(cos θ+(√3)sin θ−h)^2 +(sin θ+(√3)cos θ−k)^2 2s=((2r)/( (√((cos θ+(√3)sin θ−h)^2 +(cos θ+(√3)sin θ−h)^2 )))) say 2s=((2r)/( (√(f(θ))))) f(θ)=(cos θ+(√3)sin θ−h)^2 +(sin θ+(√3)cos θ−k)^2 f(θ)=h^2 +k^2 −2h(cos θ+(√3)sin θ) −2k(sin θ+(√3)cos θ) +(cos θ+(√3)sin θ)^2 +(sin θ+(√3)cos θ)^2 f ′(θ)=2h(sin θ−(√3)cos θ) −2k(cos θ−(√3)sin θ) +4(√3)cos 2θ =2(h+k(√3))sin θ −2(k+h(√3))cos θ +4(√3)cos 2θ let cos 2θ=s f ′(θ)=(√2)(h+k(√3))(√(2−(1+s))) −(√2)(k+h(√3))(√(1+s)) +4(√3)(1+s)−4(√3) f ′(θ)=0 ⇒ 2(h+k(√3))^2 (2−z^2 ) ={(√2)(k+h(√3))z+4(√3)z^2 −4(√3)}^2 .....](Q149988.png)
$$\:{A}\left(−\mathrm{2}{h}+\mathrm{2}{s}\mathrm{cos}\:\theta,\:−\mathrm{2}{k}\right)\:\:\:;\: \\ $$$$\:{B}\left(−\mathrm{2}{h},\:−\mathrm{2}{k}+\mathrm{2}{s}\mathrm{sin}\:\theta\right) \\ $$$$\:{side}\:{of}\:{eql}.\bigtriangleup\:=\:\mathrm{2}{s} \\ $$$$\:\underset{−} {{Eq}.\:{of}\:{circle}} \\ $$$$\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:\:\:\:....\left(\mathrm{1}\right) \\ $$$$\:\:\underset{−} {{Eq}.\:{of}\:{DP}} \\ $$$$\:\:{y}−{s}\mathrm{sin}\:\theta+{k}=\mathrm{cot}\:\theta\left({x}−{s}\mathrm{cos}\:\theta+{h}\right) \\ $$$$\:\:{using}\:{this}\:{in}\:..\left(\mathrm{1}\right) \\ $$$$\:\:{x}^{\mathrm{2}} +\left[{h}\mathrm{cot}\:\theta−{k}+{s}\mathrm{sin}\:\theta\right. \\ $$$$\left.\:\:\:\:\:\:\:+\mathrm{cot}\:\theta\left({x}−{s}\mathrm{cos}\:\theta\right)\right]^{\mathrm{2}} \\ $$$$\:\:\:={r}^{\mathrm{2}} \\ $$$$\:\:{further} \\ $$$$\:\:{x}−{s}\mathrm{cos}\:\theta+{h}={s}\sqrt{\mathrm{3}}\mathrm{sin}\:\theta \\ $$$$\:\:\Rightarrow\:\frac{{r}^{\mathrm{2}} }{{s}^{\mathrm{2}} }=\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{h}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−{k}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{s}=\frac{\mathrm{2}{r}}{\:\sqrt{\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{h}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{h}\right)^{\mathrm{2}} }} \\ $$$${say}\:\:\mathrm{2}{s}=\frac{\mathrm{2}{r}}{\:\sqrt{{f}\left(\theta\right)}} \\ $$$${f}\left(\theta\right)=\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta−{h}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−{k}\right)^{\mathrm{2}} \\ $$$${f}\left(\theta\right)={h}^{\mathrm{2}} +{k}^{\mathrm{2}} −\mathrm{2}{h}\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:−\mathrm{2}{k}\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:+\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:+\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$${f}\:'\left(\theta\right)=\mathrm{2}{h}\left(\mathrm{sin}\:\theta−\sqrt{\mathrm{3}}\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:−\mathrm{2}{k}\left(\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:+\mathrm{4}\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\:\:=\mathrm{2}\left({h}+{k}\sqrt{\mathrm{3}}\right)\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:−\mathrm{2}\left({k}+{h}\sqrt{\mathrm{3}}\right)\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:+\mathrm{4}\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta \\ $$$${let}\:\:\:\mathrm{cos}\:\mathrm{2}\theta={s} \\ $$$${f}\:'\left(\theta\right)=\sqrt{\mathrm{2}}\left({h}+{k}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{2}−\left(\mathrm{1}+{s}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\sqrt{\mathrm{2}}\left({k}+{h}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{1}+{s}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}+{s}\right)−\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${f}\:'\left(\theta\right)=\mathrm{0}\:\:\Rightarrow \\ $$$$\mathrm{2}\left({h}+{k}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{2}−{z}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\left\{\sqrt{\mathrm{2}}\left({k}+{h}\sqrt{\mathrm{3}}\right){z}+\mathrm{4}\sqrt{\mathrm{3}}{z}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\right\}^{\mathrm{2}} \\ $$$$..... \\ $$
Commented by mr W last updated on 09/Aug/21
$${great}\:{solution}! \\ $$