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Question Number 184387 by ajfour last updated on 05/Jan/23

Commented by ajfour last updated on 05/Jan/23

$I f a l l t h r e e c o l o u r e d a r e a s a r e$ $e q u a l t h e n f i n d p a r a m e t e r s$ $h a n d k o f p a r a b o l a y = {(x - h)}^{2} + k .$
	Answered by mr W last updated on 06/Jan/23

	Commented by mr W last updated on 06/Jan/23

	Commented by mr W last updated on 07/Jan/23

	$a r e a o f r e d t r i a n g l e A_{1} = \frac{2 \times 1}{2} = 1$ $e q n . o f l i n e :$ $y = \frac{x}{2} + 1$ $e q n . o f p a r a b o l a :$ $y = {(x - h)}^{2} + k$ $i n t e r s e c t i o n o f l i n e a n d p a r a b o l a :$ ${(x - h)}^{2} + k = \frac{x}{2} + 1$ ${(x - h)}^{2} - \frac{(x - h)}{2} + (k - 1 - \frac{h}{2}) = 0$ $\Rightarrow x_{2} - h = \frac{1}{4} + \frac{1}{4} \sqrt{17 + 8 h - 16 k}$ $\Rightarrow y_{2} = \frac{9}{8} + \frac{h}{2} + \frac{1}{8} \sqrt{17 + 8 h - 16 k}$ $a r e a o f b l u e r e g i o n :$ $A_{3} = \frac{1}{6} {[\frac{1}{4} - 4 (k - 1 - \frac{h}{2})]}^{3 / 2} = A_{1} = 1$ $\frac{1}{6} {[\frac{1}{4} - 4 (k - 1 - \frac{h}{2})]}^{3 / 2} = 1$ $17 - 16 k + 8 h = 4 \sqrt[3]{36}$ $\Rightarrow k = \frac{h}{2} + \frac{17}{16} - \frac{\sqrt[3]{36}}{4}$ $\Rightarrow x_{2} = h + \frac{1}{4} + \frac{\sqrt[3]{6}}{2}$ $\Rightarrow y_{2} = \frac{h}{2} + \frac{9}{8} + \frac{\sqrt[3]{6}}{4}$ $s i n c e A_{2} = A_{3}, a r e a u n d e r t h e p a r a b o l a$ $i s e q u a l t o a r e a u n d e r t h e l i n e f r o m$ $x = 0 t o x = x_{2} .$ $x_{2} k + \frac{h^{3}}{3} + \frac{{(x_{2} - h)}^{3}}{3} = \frac{(1 + y_{2}) x_{2}}{2}$ $(h + \frac{1}{4} + \frac{\sqrt[3]{6}}{2}) (\frac{h}{2} + \frac{17}{16} - \frac{\sqrt[3]{36}}{4}) + \frac{h^{3}}{3} + \frac{{(\frac{1}{4} + \frac{\sqrt[3]{6}}{2})}^{3}}{3} = \frac{(1 + \frac{h}{2} + \frac{9}{8} + \frac{\sqrt[3]{6}}{4}) (h + \frac{1}{4} + \frac{\sqrt[3]{6}}{2})}{2}$ $\Rightarrow h^{3} + \frac{3}{4} (h + \frac{1}{4} + \frac{\sqrt[3]{6}}{2}) (h - \frac{\sqrt[3]{6}}{2} - \sqrt[3]{36}) + {(\frac{1}{4} + \frac{\sqrt[3]{6}}{2})}^{3} = 0$ $s i n c e t h i s i s a c u b i c e q u a t i o n f o r h,$ $e x a c t s o l u t i o n i s p o s s i b l e, b u t v e r y$ $c o m p l i c a t e d . w e g e t a p p r o x i m a t e l y$ $h \approx 1.5671, k \approx 1.0206$
	Commented by ajfour last updated on 06/Jan/23

	$M a r v e l l o u s S i r! I s h a l l a t t e m p t t o o .$
	Answered by ajfour last updated on 07/Jan/23
	$If vertex of parabola is Origin. y=x^2 [eq. of parabola] y=(x/2)+(1+(h/2)−k) [eq. of line] say 1+(h/2)−k=m t^2 =(t/2)+m ∫_(−h) ^( t) (x^2 −(x/2)−m)dx=A (=0,1) ((t^3 /3)−(t^2 /4)+mt)+((h^3 /3)+(h^2 /4)+mh)=A ⇒ (t^3 /3)−(t^2 /4)+t^3 −(t^2 /2) +((h^3 /3)+(h^2 /4)+t^2 h−((th)/2))=A ⇒ (t+h){((t^2 −th+h^2 )/3)−(((t−h))/4) +t^2 −(t/2)}=A For A=0, t=t_2 as t+h≠0 ⇒ 4(t^2 −th+h^2 )+12t^2 +3h=6t+3t ⇒ 16t^2 −4th+4h^2 =9t−3h ⇒ 4((t^2 /h^2 )−(t/h)+1)=(3/h)(((3t)/h)−1) say (t/h)=s 4s^2 −(4+(9/h))s+(4+(3/h))=0 ⇒ s=(1/2)(1+(9/(4h)))±(√((1/4)(1+(9/(4h)))^2 −(1+(3/(4h))))) ⇒ 8hs=4h+9±(√((4h+9)^2 −16h(4h+3))) 8hs=4h+9±(√(81+24h−48h^2 )) ....$
	$I f v e r t e x o f p a r a b o l a i s O r i g i n .$ $y = x^{2} [e q . o f p a r a b o l a]$ $y = \frac{x}{2} + (1 + \frac{h}{2} - k) [e q . o f l i n e]$ $s a y 1 + \frac{h}{2} - k = m$ $t^{2} = \frac{t}{2} + m$ $\int_{- h}^{t} (x^{2} - \frac{x}{2} - m) d x = A (= 0, 1)$ $(\frac{t^{3}}{3} - \frac{t^{2}}{4} + m t) + (\frac{h^{3}}{3} + \frac{h^{2}}{4} + m h) = A$ $\Rightarrow \frac{t^{3}}{3} - \frac{t^{2}}{4} + t^{3} - \frac{t^{2}}{2}$ $+ (\frac{h^{3}}{3} + \frac{h^{2}}{4} + t^{2} h - \frac{t h}{2}) = A$ $\Rightarrow (t + h) {\frac{t^{2} - t h + h^{2}}{3} - \frac{(t - h)}{4}$ $+ t^{2} - \frac{t}{2}} = A$ $F o r A = 0, t = t_{2}$ $a s t + h \neq 0 \Rightarrow$ $4 (t^{2} - t h + h^{2}) + 12 t^{2} + 3 h = 6 t + 3 t$ $\Rightarrow 16 t^{2} - 4 t h + 4 h^{2} = 9 t - 3 h$ $\Rightarrow 4 (\frac{t^{2}}{h^{2}} - \frac{t}{h} + 1) = \frac{3}{h} (\frac{3 t}{h} - 1)$ $s a y \frac{t}{h} = s$ $4 s^{2} - (4 + \frac{9}{h}) s + (4 + \frac{3}{h}) = 0$ $\Rightarrow s = \frac{1}{2} (1 + \frac{9}{4 h}) \pm \sqrt{\frac{1}{4} {(1 + \frac{9}{4 h})}^{2} - (1 + \frac{3}{4 h})}$ $\Rightarrow 8 h s = 4 h + 9 \pm \sqrt{{(4 h + 9)}^{2} - 16 h (4 h + 3)}$ $8 h s = 4 h + 9 \pm \sqrt{81 + 24 h - 48 h^{2}}$ $. . . .$
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