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Question Number 43386 by ajfour last updated on 10/Sep/18

Commented by ajfour last updated on 10/Sep/18

$I f t h e t w o d i f f e r e n t l y s h a d e d$ $a r e a s a r e e q u a l, f i n d s l o p e m o f$ $l i n e, i n t e r m s o f a .$
Commented by Tawa1 last updated on 10/Sep/18

$Sir, please how did you draw the smoth curve using lekh diagram ?$
Commented by ajfour last updated on 10/Sep/18

$q u i c k l y m o v e d f i n g e r l i k e a$ $p a r a b o l a (o n e b r a n c h l o n g, o t h e r$ $s h o r t), (l e k h r e c o g n i s e d i t),$ $c o p i e d i t, f l i p p e d t h e c o p y$ $v e r i c a l t h e n h o r i z o n t a l a n d$ $j o i n e d t h e b r a n c h e n d s o f$ $o r i g i n a l a n d c o p i e d o n e, t h e n$ $g r o u p e d t h e m, t h e n r e s i z e d i t .$
Commented by Tawa1 last updated on 10/Sep/18

$God bless you sir$
	Answered by MrW3 last updated on 10/Sep/18
	$y=x(x^2 −a^2 ) (dy/dx)=3x^2 −a^2 at x=0: tan θ_0 =−a^2 ⇒r sin θ=r cos θ(r^2 cos^2 θ−a^2 ) ⇒r^2 =((tan θ+a^2 )/(cos^2 θ)) A=∫_θ_0 ^(2π) ((r^2 dθ)/2)=(1/2)∫_θ_0 ^(2π) ((tan θ+a^2 )/(cos^2 θ)) dθ =(1/2)∫_θ_0 ^(2π) (tan θ+a^2 ) d(tan θ) =(1/2)[((tan^2 θ)/2)+a^2 tan θ]_θ_0 ^(2π) =(1/2)[−(a^4 /2)+a^4 ]=(a^4 /4) let tan θ_1 =m A_1 =(1/2)[((tan^2 θ)/2)+a^2 tan θ]_θ_1 ^(2π) =(1/2)[−((tan^2 θ_1 )/2)−a^2 tan θ_1 ] =−(1/2)[(m^2 /2)+a^2 m]=(A/2)=(a^4 /8) ⇒2m^2 +4a^2 m+a^4 =0 ⇒m=((−4a^2 ±(√(16a^4 −8a^4 )))/4) = { ((−((2+(√2))/2)a^2 )),((−((2−(√2))/2)a^2 )) :} since ∣θ_1 ∣<∣θ_0 ∣⇒∣m∣=∣tan θ_1 ∣<∣tan θ_0 ∣=a^2 the suitable solution is: m=−((2−(√2))/2)a^2 ≈−0.293a^2$
	$y = x (x^{2} - a^{2})$ $\frac{d y}{d x} = 3 x^{2} - a^{2}$ $a t x = 0 : \tan θ_{0} = - a^{2}$ $\Rightarrow r \sin θ = r \cos θ (r^{2} \cos^{2} θ - a^{2})$ $\Rightarrow r^{2} = \frac{\tan θ + a^{2}}{\cos^{2} θ}$ $A = \int_{θ_{0}}^{2 π} \frac{r^{2} d θ}{2} = \frac{1}{2} \int_{θ_{0}}^{2 π} \frac{\tan θ + a^{2}}{\cos^{2} θ} d θ$ $= \frac{1}{2} \int_{θ_{0}}^{2 π} (\tan θ + a^{2}) d (\tan θ)$ $= \frac{1}{2} {[\frac{\tan^{2} θ}{2} + a^{2} \tan θ]}_{θ_{0}}^{2 π}$ $= \frac{1}{2} [- \frac{a^{4}}{2} + a^{4}] = \frac{a^{4}}{4}$ $l e t \tan θ_{1} = m$ $A_{1} = \frac{1}{2} {[\frac{\tan^{2} θ}{2} + a^{2} \tan θ]}_{θ_{1}}^{2 π}$ $= \frac{1}{2} [- \frac{\tan^{2} θ_{1}}{2} - a^{2} \tan θ_{1}]$ $= - \frac{1}{2} [\frac{m^{2}}{2} + a^{2} m] = \frac{A}{2} = \frac{a^{4}}{8}$ $\Rightarrow 2 m^{2} + 4 a^{2} m + a^{4} = 0$ $\Rightarrow m = \frac{- 4 a^{2} \pm \sqrt{16 a^{4} - 8 a^{4}}}{4}$ $= {\begin{cases} - \frac{2 + \sqrt{2}}{2} a^{2} \\ - \frac{2 - \sqrt{2}}{2} a^{2} \end{cases}$ $s i n c e ∣ θ_{1} ∣<∣ θ_{0} ∣\Rightarrow∣ m ∣=∣ \tan θ_{1} ∣<∣ \tan θ_{0} ∣= a^{2}$ $t h e s u i t a b l e s o l u t i o n i s :$ $m = - \frac{2 - \sqrt{2}}{2} a^{2} \approx - 0.293 a^{2}$
	Commented by ajfour last updated on 10/Sep/18

	$T h a n k s S i r, i g o t t h e s a m e a n s w e r .$
	Answered by ajfour last updated on 10/Sep/18

	$I n t e r s e c t i o n p o i n t s o f l i n e$ $a n d c u r v e :$ $y = m x = x (x^{2} - a^{2})$ $\Rightarrow O (0, 0) a n d P (\sqrt{a^{2} + m}, y_{P})$ $A = - \int_{0}^{a} (x^{3} - a^{2} x) d x = - (\frac{a^{4}}{4} - \frac{a^{4}}{2})$ $\Rightarrow A = \frac{a^{4}}{4}$ $\int_{0}^{\sqrt{a^{2} + m}} (m x - x^{3} + a^{2} x) d x = \frac{a^{4}}{8} (= \frac{A}{2})$ $\Rightarrow \frac{m (a^{2} + m)}{2} - \frac{{(a^{2} + m)}^{2}}{4} + \frac{a^{2} (a^{2} + m)}{2} = \frac{a^{4}}{8}$ $\Rightarrow {(a^{2} + m)}^{2} = \frac{a^{4}}{2}$ $\Rightarrow m = (\frac{1}{\sqrt{2}} - 1) a^{2}$ $s a m e a s m = - (\frac{2 - \sqrt{2}}{2}) a^{2} .$ $[\frac{d y}{d x} ∣_{x = 0} = - a^{2}; s o w e r e j e c t$ $m = - (1 + \frac{1}{\sqrt{2}}) a^{2}] .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
	$solving mx=x(x^2 −a^2 ) x{m−(x^2 −a^2 }=0 x=0 x^2 −a^2 =m x=±(√(a^2 +m)) y=mx intersect the curve at {−(√(a^2 +m)) ,0,(√(a^2 +m)) } the curve y=x(x^2 −a^2 ) intersect x axis at{−a,0,a} now area of dot shaded+line shaded= ∫_0 ^a x(x^2 −a^2 )dx =∫_0 ^a x^3 −x^ a^2 dx =∣(x^4 /4)−(x^2 /2)×a^2 ∣_0 ^a =(a^4 /4)−(a^4 /2) taking the mod of value of area=(a^4 /4) so dot shaded=(a^4 /8) line shaded=(a^4 /8) area line shaded ∫_0 ^((√(a^2 +m)) ) mx dx+∫_((√(a^2 +m)) ) ^a x(x^2 −a^2 )dx =∣((mx^2 )/2)∣_0 ^((√(a^2 +m)) ) +∣(x^4 /4)−((x^2 a^2 )/2)∣_((√(a^2 +m)) ) ^a =((m(a^2 +m))/2)+{((a^4 /4)−(a^4 /2))−(((a^4 +2a^2 m+m^2 )/4)−((a^2 (a^2 +m))/2)} =((a^2 m+m^2 )/2)+{(((−a^4 )/4))−(((a^4 +2a^2 m+m^2 −2a^4 −2a^2 m)/4))} =((a^2 m+m^2 )/2)+{((−a^4 −a^4 −2a^2 m−m^2 +2a^4 +2a^2 m)/4)} =((a^2 m+m^2 )/2)−(m^2 /4) =((2a^2 m+2m^2 −m^2 )/4) =((2a^2 m+m^2 )/4) ((2a^2 m+m^2 )/4)=(a^4 /8) 4a^2 m+2m^2 −a^4 =0 2m^2 +4a^2 m−a^4 =0 m=((−4a^2 ±(√(16a^4 +8a^4 )) )/4) =((−4a^2 ±2(√6) a^2 )/4)=((−2a^2 ±(√6) a^2 )/2) as per figure m is negdtive m=((−a^2 (2+(√6) ))/2) pls check$
	$s o l v i n g m x = x (x^{2} - a^{2})$ $x {m - (x^{2} - a^{2}} = 0$ $x = 0$ $x^{2} - a^{2} = m$ $x = \pm \sqrt{a^{2} + m}$ $y = m x i n t e r s e c t t h e c u r v e a t {- \sqrt{a^{2} + m}, 0, \sqrt{a^{2} + m}}$ $t h e c u r v e y = x (x^{2} - a^{2}) i n t e r s e c t x a x i s a t {- a, 0, a}$ $n o w a r e a o f d o t s h a d e d + l i n e s h a d e d =$ $\int_{0}^{a} x (x^{2} - a^{2}) d x$ $= \int_{0}^{a} x^{3} - x^{} a^{2} d x$ $=∣ \frac{x^{4}}{4} - \frac{x^{2}}{2} \times a^{2} ∣_{0}^{a}$ $= \frac{a^{4}}{4} - \frac{a^{4}}{2}$ $t a k i n g t h e m o d o f v a l u e o f a r e a = \frac{a^{4}}{4}$ $s o d o t s h a d e d = \frac{a^{4}}{8}$ $l i n e s h a d e d = \frac{a^{4}}{8}$ $a r e a l i n e s h a d e d$ $\int_{0}^{\sqrt{a^{2} + m}} m x d x + \int_{\sqrt{a^{2} + m}}^{a} x (x^{2} - a^{2}) d x$ $=∣ \frac{{m x}^{2}}{2} ∣_{0}^{\sqrt{a^{2} + m}} + ∣ \frac{x^{4}}{4} - \frac{x^{2} a^{2}}{2} ∣_{\sqrt{a^{2} + m}}^{a}$ $= \frac{m (a^{2} + m)}{2} + {(\frac{a^{4}}{4} - \frac{a^{4}}{2}) - (\frac{a^{4} + 2 a^{2} m + m^{2}}{4} - \frac{a^{2} (a^{2} + m)}{2}}$ $= \frac{a^{2} m + m^{2}}{2} + {(\frac{- a^{4}}{4}) - (\frac{a^{4} + 2 a^{2} m + m^{2} - 2 a^{4} - 2 a^{2} m}{4})}$ $= \frac{a^{2} m + m^{2}}{2} + {\frac{- a^{4} - a^{4} - 2 a^{2} m - m^{2} + 2 a^{4} + 2 a^{2} m}{4}}$ $= \frac{a^{2} m + m^{2}}{2} - \frac{m^{2}}{4}$ $= \frac{2 a^{2} m + 2 m^{2} - m^{2}}{4}$ $= \frac{2 a^{2} m + m^{2}}{4}$ $\frac{2 a^{2} m + m^{2}}{4} = \frac{a^{4}}{8}$ $4 a^{2} m + 2 m^{2} - a^{4} = 0$ $2 m^{2} + 4 a^{2} m - a^{4} = 0$ $m = \frac{- 4 a^{2} \pm \sqrt{16 a^{4} + 8 a^{4}}}{4}$ $= \frac{- 4 a^{2} \pm 2 \sqrt{6} a^{2}}{4} = \frac{- 2 a^{2} \pm \sqrt{6} a^{2}}{2}$ $a s p e r f i g u r e m i s n e g d t i v e$ $m = \frac{- a^{2} (2 + \sqrt{6})}{2}$ $p l s c h e c k$
	Commented by ajfour last updated on 10/Sep/18

	$p l e a s e c h e c k, s o m e l i t t l e e r r o r, S i r .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$i h a v e d o n e$ $\int_{0}^{\sqrt{a^{2} + m}} m x + \int_{\sqrt{a^{2} + m}}^{a} x (x^{2} - a^{2}) d x t o f i n d t h e l i n e s h a d e d$ $a r e a$ $s u m o f t h e v a l u e o f t h i s t w o i n t r e g a l = \frac{a^{4}}{8}$ $b u t y o u h a v e d o n e$ $\int_{0}^{\sqrt{a^{2} + m}} m x - x (x^{2} - a^{2}) d x = \frac{a^{4}}{8}$ $p l s s a y w h i c h a r e a b e l o n g s t o t h i s i n t r e g a l$
	Commented by ajfour last updated on 10/Sep/18

	$t h e d o t s h a d e d a r e a i s e q u a l t o$ $w h a t i h a v e i n t e g r a t e d .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$i d o n o t a g r e e . . . \int_{0}^{\sqrt{a^{2} + m}} m x d x i s t h e a r e a o f t r i a n g l e$ $f o r m e d b y y = m x a n d x a x i s a n d s t l i n e x = \sqrt{a^{2} + m}$ $\int_{0}^{\sqrt{a^{2} + m}} x (x^{2} - a^{2}) d x$ $i s t h e a r e a f o r m e d b y x a x i s a n d c u r v e x (x^{2} - a^{2}) a n d s t$ $l i n e x = \sqrt{a^{2} + m} [$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$p l s f i n d l i n e s h a d e d a r e a . . .$
	Commented by ajfour last updated on 10/Sep/18

	$h o w w i l l y o u f i n d a r e a o f$ ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}, S i r ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$y o u f i n d e r r o r i n m y c a l c u l a t i o n . . . l o c a t e i t$
	Commented by ajfour last updated on 10/Sep/18

	$\int_{0}^{\sqrt{a^{2} + m}} m x d x + \int_{\sqrt{a^{2} + m}}^{a} x (x^{2} - a^{2}) d x = - \frac{a^{4}}{8}$ $(a r e a i s b e l o w x - a x i s s o - v e)$ $t h e n y o u s h a l l g e t$ $4 a^{2} m + 2 m^{2} + a^{4} = 0$ $\Rightarrow 2 m^{2} + 4 a^{2} m + a^{4} = 0$ $m = \frac{- 4 a^{2} \pm \sqrt{16 a^{4} - 8 a^{4}}}{4}$ $o r m = (- 1 + \frac{1}{\sqrt{2}}) a^{2}$ $(n o t a c c e p t i n g t h e o t h e r r o o t) .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

	$o k t h a n k y o u . . .$
	Answered by MJS last updated on 10/Sep/18

	$p : y = x^{3} - a^{2} x$ $l : y = m x$ $p \cap l$ $x^{3} - (m + a^{2}) x = 0 \Rightarrow x = \sqrt{m + a^{2}}$ $dotted area : D = \underset{0}{\int^{\sqrt{m + a^{2}}}} (x^{3} - a^{2} x) d x - m \underset{0}{\int^{\sqrt{m + a^{2}}}} x d x$ $shaded area : S = m \underset{0}{\int^{\sqrt{m + a^{2}}}} x d x + \underset{\sqrt{m + a^{2}}}{\int^{a}} (x^{3} - a^{2} x) d x$ $D = \frac{1}{4} (m^{2} - a^{4}) - \frac{1}{2} m (m + a^{2}) = - \frac{1}{4} {(m + a^{2})}^{2}$ $S = \frac{1}{2} m (m + a^{2}) - \frac{1}{4} m^{2} = \frac{1}{4} m (m + 2 a^{2})$ $S - D = 0$ $m^{2} + 2 a^{2} m + \frac{a^{4}}{2} = 0$ $m = (- 1 + \frac{\sqrt{2}}{2}) a^{2} [the 2^{nd} solution {doesn}^{'} t$ $give a real value for \sqrt{m + a^{2}}]$
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