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Question Number 149940 by ajfour last updated on 08/Aug/21

Commented by ajfour last updated on 08/Aug/21

$s e e Q . 149894$
	Answered by ajfour last updated on 09/Aug/21
	$A(−2h+2scos θ, −2k) ; B(−2h, −2k+2ssin θ) side of eql.△ = 2s Eq. of circle_(−) x^2 +y^2 =r^2 ....(1) Eq. of DP_(−) y−ssin θ+k=cot θ(x−scos θ+h) using this in ..(1) x^2 +[hcot θ−k+ssin θ +cot θ(x−scos θ)]^2 =r^2 further x−scos θ+h=s(√3)sin θ ⇒ (r^2 /s^2 )=(cos θ+(√3)sin θ−h)^2 +(sin θ+(√3)cos θ−k)^2 2s=((2r)/( (√((cos θ+(√3)sin θ−h)^2 +(cos θ+(√3)sin θ−h)^2 )))) say 2s=((2r)/( (√(f(θ))))) f(θ)=(cos θ+(√3)sin θ−h)^2 +(sin θ+(√3)cos θ−k)^2 f(θ)=h^2 +k^2 −2h(cos θ+(√3)sin θ) −2k(sin θ+(√3)cos θ) +(cos θ+(√3)sin θ)^2 +(sin θ+(√3)cos θ)^2 f ′(θ)=2h(sin θ−(√3)cos θ) −2k(cos θ−(√3)sin θ) +4(√3)cos 2θ =2(h+k(√3))sin θ −2(k+h(√3))cos θ +4(√3)cos 2θ let cos 2θ=s f ′(θ)=(√2)(h+k(√3))(√(2−(1+s))) −(√2)(k+h(√3))(√(1+s)) +4(√3)(1+s)−4(√3) f ′(θ)=0 ⇒ 2(h+k(√3))^2 (2−z^2 ) ={(√2)(k+h(√3))z+4(√3)z^2 −4(√3)}^2 .....$
	$A (- 2 h + 2 s \cos θ, - 2 k);$ $B (- 2 h, - 2 k + 2 s \sin θ)$ $s i d e o f e q l . △ = 2 s$ $\underline{E q . o f c i r c l e}$ $x^{2} + y^{2} = r^{2} . . . . (1)$ $\underline{E q . o f D P}$ $y - s \sin θ + k = \cot θ (x - s \cos θ + h)$ $u s i n g t h i s i n . . (1)$ $x^{2} + [h \cot θ - k + s \sin θ$ ${+ \cot θ (x - s \cos θ)]}^{2}$ $= r^{2}$ $f u r t h e r$ $x - s \cos θ + h = s \sqrt{3} \sin θ$ $\Rightarrow \frac{r^{2}}{s^{2}} = {(\cos θ + \sqrt{3} \sin θ - h)}^{2}$ $+ {(\sin θ + \sqrt{3} \cos θ - k)}^{2}$ $2 s = \frac{2 r}{\sqrt{{(\cos θ + \sqrt{3} \sin θ - h)}^{2} + {(\cos θ + \sqrt{3} \sin θ - h)}^{2}}}$ $s a y 2 s = \frac{2 r}{\sqrt{f (θ)}}$ $f (θ) = {(\cos θ + \sqrt{3} \sin θ - h)}^{2}$ $+ {(\sin θ + \sqrt{3} \cos θ - k)}^{2}$ $f (θ) = h^{2} + k^{2} - 2 h (\cos θ + \sqrt{3} \sin θ)$ $- 2 k (\sin θ + \sqrt{3} \cos θ)$ $+ {(\cos θ + \sqrt{3} \sin θ)}^{2}$ $+ {(\sin θ + \sqrt{3} \cos θ)}^{2}$ $f^{'} (θ) = 2 h (\sin θ - \sqrt{3} \cos θ)$ $- 2 k (\cos θ - \sqrt{3} \sin θ)$ $+ 4 \sqrt{3} \cos 2 θ$ $= 2 (h + k \sqrt{3}) \sin θ$ $- 2 (k + h \sqrt{3}) \cos θ$ $+ 4 \sqrt{3} \cos 2 θ$ $l e t \cos 2 θ = s$ $f^{'} (θ) = \sqrt{2} (h + k \sqrt{3}) \sqrt{2 - (1 + s)}$ $- \sqrt{2} (k + h \sqrt{3}) \sqrt{1 + s}$ $+ 4 \sqrt{3} (1 + s) - 4 \sqrt{3}$ $f^{'} (θ) = 0 \Rightarrow$ $2 {(h + k \sqrt{3})}^{2} (2 - z^{2})$ $= {\sqrt{2} (k + h \sqrt{3}) z + 4 \sqrt{3} z^{2} - 4 \sqrt{3}}^{2}$ $. . . . .$
	Commented by mr W last updated on 09/Aug/21

	$g r e a t s o l u t i o n!$
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