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Question Number 80550 by ajfour last updated on 04/Feb/20

Commented by ajfour last updated on 04/Feb/20

$I f b o t h c o l o u r e d a r e a s a r e$ $e q u a l, f i n d a .$
	Answered by mr W last updated on 04/Feb/20

	Commented by jagoll last updated on 04/Feb/20

	$w a w . . . i n t e g r a l s e s i o n$
	Commented by ajfour last updated on 04/Feb/20

	$y e s s i r, y o u r s o l u t i o n i s p e r f e c t!$
	Commented by mr W last updated on 04/Feb/20

	$i f o u n d s o m e t y p o s, n o w f i x e d .$ $i t h i n k y o u r e q u a t i o n a r r i v e s a t t h e$ $s a m e r e s u l t .$
	Commented by john santu last updated on 04/Feb/20

	$l e t P (c, c^{2})$ $a r e a b l u e = \frac{1}{2} (a - c) . c^{2} + \frac{1}{3} c^{3}$ $= \frac{1}{2} {a c}^{2} - \frac{1}{6} c^{3} .$ $l e t Q (- b, b^{2}) . l i n e P Q : y = - (\frac{b^{2}}{a + b}) (x - a)$ $a r e a g r e e n = \int_{- b}^{c} - (\frac{b^{2}}{a + b}) (x - a) - x^{2} d x$
	Commented by ajfour last updated on 04/Feb/20

	$l o o k s p r o m i s i n g s i r!$
	Commented by mr W last updated on 04/Feb/20
	$P(p,p^2 ), Q(q,q^2 ), R(a,0) y′=2x=2p eqn. of normal to parabola at P: y=p^2 −(1/(2p))(x−p) point Q: q^2 =p^2 −(1/(2p))(q−p) 2pq^2 +q−p(2p^2 +1)=0 ⇒q=((−1±(4p^2 +1))/(4p))= { ((p ⇒no solution)),((−((1+2p^2 )/(2p))=−(1/(2p))−p)) :} point R: 0=p^2 −(1/(2p))(a−p) ⇒a=p(2p^2 +1) A_(blue) =(p^3 /3)+(((a−p)p^2 )/2)=((3ap^2 −p^3 )/6) A_(green) =(((p−q)(p^2 +q^2 ))/2)−(p^3 /3)+(q^3 /3) A_(green) =((p^3 −3p^2 q+3pq^2 −q^3 )/6) A_(blue) =A_(green) 3ap^2 −p^3 =p^3 −3p^2 q+3pq^2 −q^3 2p^3 −3(a+q)p^2 +(3p−q)q^2 =0 48p^8 −48p^6 −48p^4 −12p^2 −1=0 with t=p^2 >0 ⇒48t^4 −48t^3 −48t^2 −12t−1=0 t=(1/4)(1+(√3)+(√((42+8(√3))/3)))≈1.761750 ⇒p=(√t)≈1.327309 ⇒a=p(2p^2 +1)≈6.004086$
	$P (p, p^{2}), Q (q, q^{2}), R (a, 0)$ $y^{'} = 2 x = 2 p$ $e q n . o f n o r m a l t o p a r a b o l a a t P :$ $y = p^{2} - \frac{1}{2 p} (x - p)$ $p o i n t Q :$ $q^{2} = p^{2} - \frac{1}{2 p} (q - p)$ $2 {p q}^{2} + q - p (2 p^{2} + 1) = 0$ $\Rightarrow q = \frac{- 1 \pm (4 p^{2} + 1)}{4 p} = {\begin{cases} p \Rightarrow n o s o l u t i o n \\ - \frac{1 + 2 p^{2}}{2 p} = - \frac{1}{2 p} - p \end{cases}$ $p o i n t R :$ $0 = p^{2} - \frac{1}{2 p} (a - p)$ $\Rightarrow a = p (2 p^{2} + 1)$ $A_{b l u e} = \frac{p^{3}}{3} + \frac{(a - p) p^{2}}{2} = \frac{3 {a p}^{2} - p^{3}}{6}$ $A_{g r e e n} = \frac{(p - q) (p^{2} + q^{2})}{2} - \frac{p^{3}}{3} + \frac{q^{3}}{3}$ $A_{g r e e n} = \frac{p^{3} - 3 p^{2} q + 3 {p q}^{2} - q^{3}}{6}$ $A_{b l u e} = A_{g r e e n}$ $3 {a p}^{2} - p^{3} = p^{3} - 3 p^{2} q + 3 {p q}^{2} - q^{3}$ $2 p^{3} - 3 (a + q) p^{2} + (3 p - q) q^{2} = 0$ $48 p^{8} - 48 p^{6} - 48 p^{4} - 12 p^{2} - 1 = 0$ $w i t h t = p^{2} > 0$ $\Rightarrow 48 t^{4} - 48 t^{3} - 48 t^{2} - 12 t - 1 = 0$ $t = \frac{1}{4} (1 + \sqrt{3} + \sqrt{\frac{42 + 8 \sqrt{3}}{3}}) \approx 1.761750$ $\Rightarrow p = \sqrt{t} \approx 1.327309$ $\Rightarrow a = p (2 p^{2} + 1) \approx 6.004086$
	Commented by mr W last updated on 04/Feb/20

	Commented by TawaTawa last updated on 04/Feb/20

	$Weldone sir . God bless you more .$
	Commented by ajfour last updated on 04/Feb/20

	$T h a n k s s i r, i b e l i e v e y o u r a n s w e r,$ $i s t h e c o r r e c t o n e . L e t m e$ $c h e c k m i n e . .$
	Commented by ajfour last updated on 04/Feb/20

	$g o a h e a d s a n t u s i r . .$
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