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Question Number 166252 by mr W last updated on 16/Feb/22

Commented by mr W last updated on 16/Feb/22

$f i n d t h e s i d e l e n g t h s o f t h e m a x i m u m$ $e q u i l a t e r a l t r i a n g l e i n s c r i b e d$ $b e t w e e n t h e p a r a b o l a s y = x^{2} a n d x = y^{2} .$
	Answered by mr W last updated on 18/Feb/22

	Commented by mr W last updated on 19/Feb/22

	$s a y e q n . o f A B i s y = m x + c$ $x^{2} - m x - c = 0$ $x_{A} + x_{B} = m$ $x_{A} x_{B} = - c$ $s^{2} = {(x_{B} - x_{A})}^{2} + {(y_{B} - y_{A})}^{2}$ $s^{2} = {(x_{B} - x_{A})}^{2} + {(x_{B} - x_{A})}^{2} {(x_{B} + x_{A})}^{2}$ $s^{2} = {(x_{B} - x_{A})}^{2} [1 + {(x_{B} + x_{A})}^{2}]$ $s^{2} = Φ = (m^{2} + 4 c) (1 + m^{2})$ $\Rightarrow c = \frac{Φ}{4 (1 + m^{2})} - \frac{m^{2}}{4} ✓$ $x_{D} = \frac{x_{A} + x_{B}}{2} = \frac{m}{2}$ $y_{D} = \frac{y_{A} + y_{B}}{2} = \frac{x_{A}^{2} + x_{B}^{2}}{2} = \frac{{(x_{A} + x_{B})}^{2} - 2 x_{A} x_{B}}{2}$ $y_{D} = \frac{m^{2}}{2} + c$ $e q n . o f D C :$ $y = y_{D} - \frac{1}{m} (x - x_{D})$ $y = \frac{m^{2}}{2} + c - \frac{1}{m} (x - \frac{m}{2})$ $y = \frac{1 + m^{2}}{2} + c - \frac{x}{m}$ $y^{2} + m y - \frac{m (1 + m^{2})}{2} - m c = 0$ $y_{C} = \frac{- m + \sqrt{m (2 m^{2} + m + 2 + 4 c)}}{2}$ ${m x}_{C} - y_{C} + c = (1 + m^{2}) (\frac{m^{2}}{2} + c - y_{C})$ ${m x}_{C} - y_{C} + c = (1 + m^{2}) (\frac{m^{2} + m}{2} + c - \frac{\sqrt{m (2 m^{2} + m + 2 + 4 c)}}{2})$ $C D = \frac{\sqrt{3} s}{2}$ $\frac{3 s^{2}}{4} = \frac{{({m x}_{C} - y_{C} + c)}^{2}}{1 + m^{2}}$ $3 Φ = (1 + m^{2}) {[m^{2} + m + 2 c - \sqrt{m (2 m^{2} + m + 2 + 4 c)}]}^{2}$ $Φ_{m a x} \approx 0.1541 a t m \approx 1.018$ $s_{m a x} \approx 0.3926$
	Commented by mr W last updated on 19/Feb/22

	Commented by mr W last updated on 19/Feb/22

	Commented by ajfour last updated on 19/Feb/22
	$A(p,p^2 ) C[p+scos (θ+60°), p^2 +ssin (θ+60°)] B[p+scos θ, p^2 +ssin θ] p^2 +ssin θ=(p+scos θ)^2 ⇒ sin θ−2pcos θ=scos^2 θ ⇒ 2p=tan θ−scos θ & p+scos (θ+60°) ={p^2 +ssin (θ+60°)}^2 ⇒ 8(tan θ−scos θ)+16scos (θ+60°) ={(tan θ−scos θ)^2 +4ssin (θ+60°)}^2 s_(max) ≈ 0.39254$
	$A (p, p^{2})$ $C [p + s \cos (θ + 60 °), p^{2} + s \sin (θ + 60 °)]$ $B [p + s \cos θ, p^{2} + s \sin θ]$ $p^{2} + s \sin θ = {(p + s \cos θ)}^{2}$ $\Rightarrow \sin θ - 2 p \cos θ = s \cos^{2} θ$ $\Rightarrow 2 p = \tan θ - s \cos θ$ $& p + s \cos (θ + 60 °)$ $= {p^{2} + s \sin (θ + 60 °)}^{2}$ $\Rightarrow 8 (\tan θ - s \cos θ) + 16 s \cos (θ + 60 °)$ $= {{(\tan θ - s \cos θ)}^{2} + 4 s \sin (θ + 60 °)}^{2}$ $s_{m a x} \approx 0.39254$
	Commented by ajfour last updated on 19/Feb/22

	Commented by mr W last updated on 19/Feb/22

	$n i c e s o l u t i o n! t h a n k s s i r!$
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