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Question Number 199101 by cortano12 last updated on 28/Oct/23

	Answered by ajfour last updated on 28/Oct/23

	Commented by ajfour last updated on 28/Oct/23
	$Triangle side be 2(√3)s. Rcos 30°=s(√3) ⇒ R=2 Centre C(0,1) x^2 +(y−1)^2 =4 If M be bottom end of blue line, then M(−4cos θ, 3−4sin θ) p^2 +(3+q)^2 =20 −p=−4cos θ+2sin θ −q=3−4sin θ−2cos θ paeabola axis y+q=((sin θ)/(cos θ))(x+p) for y=0 , x_A =((qcos θ)/(sin θ))−p Line through (−p,−q) ⊥ to axis y+q+((cos θ)/(sin θ))(x+p)=0 for y=0 x_D =−p−((qsin θ)/(cos θ)) now (∓s(√3)−x_D )cos θ=(x_A ±s(√3))^2 sin^2 θ say ±s(√3) =t (t+p+((qsin θ)/(cos θ)))cos θ=(((qcos θ)/(sin θ))−p−t)^2 sin^2 θ sum of roots=0 ⇒ cos θ=2sin θ(psin θ−qcos θ) ⇒ cos θ=2sin θ{3cos θ−4sin θcos θ−2cos^2 θ +4sin θcos θ−2sin^2 θ} ⇒ cos θ=2sin θ(3cos θ−2) cos^2 θ=4(1−cos^2 θ)(3cos θ−2)^2 z^2 =4(1−z^2 )(3z−2)^2 .....$
	$T r i a n g l e s i d e b e 2 \sqrt{3} s .$ $R \cos 30 ° = s \sqrt{3} \Rightarrow R = 2$ $C e n t r e C (0, 1)$ $x^{2} + {(y - 1)}^{2} = 4$ $I f M b e b o t t o m e n d o f b l u e l i n e,$ $t h e n M (- 4 \cos θ, 3 - 4 \sin θ)$ $p^{2} + {(3 + q)}^{2} = 20$ $- p = - 4 \cos θ + 2 \sin θ$ $- q = 3 - 4 \sin θ - 2 \cos θ$ $p a e a b o l a a x i s$ $y + q = \frac{\sin θ}{\cos θ} (x + p)$ $f o r y = 0, x_{A} = \frac{q \cos θ}{\sin θ} - p$ $L i n e t h r o u g h (- p, - q) ⊥ t o a x i s$ $y + q + \frac{\cos θ}{\sin θ} (x + p) = 0$ $f o r y = 0 x_{D} = - p - \frac{q \sin θ}{\cos θ}$ $n o w (\mp s \sqrt{3} - x_{D}) \cos θ = {(x_{A} \pm s \sqrt{3})}^{2} \sin^{2} θ$ $s a y \pm s \sqrt{3} = t$ $(t + p + \frac{q \sin θ}{\cos θ}) \cos θ = {(\frac{q \cos θ}{\sin θ} - p - t)}^{2} \sin^{2} θ$ $s u m o f r o o t s = 0 \Rightarrow$ $\cos θ = 2 \sin θ (p \sin θ - q \cos θ)$ $\Rightarrow \cos θ = 2 \sin θ {3 \cos θ - 4 \sin θ \cos θ - 2 \cos^{2} θ$ $+ 4 \sin θ \cos θ - 2 \sin^{2} θ}$ $\Rightarrow \cos θ = 2 \sin θ (3 \cos θ - 2)$ $\cos^{2} θ = 4 (1 - \cos^{2} θ) {(3 \cos θ - 2)}^{2}$ $z^{2} = 4 (1 - z^{2}) {(3 z - 2)}^{2}$ $. . . . .$
	Answered by mr W last updated on 28/Oct/23

	$y = x^{2}$ $l i n e 1 : y = 4 + (x - 2) \tan α$ $l i n e 2 : y = 4 + (x - 2) \tan (α + \frac{π}{3})$ $w i t h m_{1} = \tan α = k$ $m_{2} = \tan (α + \frac{π}{3}) = \frac{k + \sqrt{3}}{1 - \sqrt{3} k}$ $i n t e r s e c t i o n l i n e 1 :$ $y_{1} = x_{1}^{2} = 4 + (x_{1} - 2) m_{1}$ $\Rightarrow x_{1}^{2} - m_{1} x_{1} + 2 m_{1} - 4 = 0$ $\Rightarrow x_{1} = m_{1} - 2$ $i n t e r s e c t i o n l i n e 2 :$ $y_{2} = x_{2}^{2} = 4 + (x_{2} - 2) m_{2}$ $\Rightarrow x_{2} = m_{2} - 2$ $s^{2} = {(x_{1} - 2)}^{2} + {(y_{1} - 4)}^{2} = {(m_{1} - 4)}^{2} (m_{1}^{2} + 1)$ $s^{2} = {(x_{2} - 2)}^{2} + {(y_{2} - 4)}^{2} = {(m_{2} - 4)}^{2} (m_{2}^{2} + 1)$ ${(m_{1} - 4)}^{2} (m_{1}^{2} + 1) = {(m_{2} - 4)}^{2} (m_{2}^{2} + 1)$ $w e g e t 4 r o o t s . t h e y s h o w 2 s u i t a b l e$ $s o l u t i o n s .$
	Commented by mr W last updated on 28/Oct/23

	Commented by cortano12 last updated on 28/Oct/23

	$great sir$
	Answered by mr W last updated on 28/Oct/23

	$\underset{―}{A l t e r n a t i v e m e t h o d :}$ $a c c . t o t h e s o l u t i o n f r o m a j f o u r s i r$ $i n Q 62938, f o r a g i v e n s i d e l e n g t h s$ $o f t h e e q u i l a t e r a l t r i a n g l e, i t s$ $v e r t e x e s s a t i s f y f o l l o w i n g e q u a t i o n :$ $x^{3} - 3 {h x}^{2} + p x + q = 0$ $w i t h h = \frac{\sqrt{s^{2} - 12}}{4 \sqrt{3}}, k = \frac{3 s^{2}}{16} - \frac{1}{4},$ $p = \frac{9 h^{2} - 3 k}{2}, q = \frac{h^{2} + k^{2} - s^{2} / 3}{3 h}$ $n o w i t i s g i v e n x = 2, r e v e r s e l y w e$ $c a n g e t s .$
	Commented by mr W last updated on 28/Oct/23

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