Suppose 0<b≤a. Show that the area of intersection E∩F of the two regions defined by E={(x,y):(x^2 /a^2 )+(y^2 /b^2 )≤1} and F={(x,y):(x^2 /b^2 )+(y^2 /a^2 )≤1} is 4absin^(−1) ((b/(√(a^2 +b^2 )))).


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Question Number 2004 by Yozzi last updated on 29/Oct/15
$Suppose 0<b≤a. Show that the area of intersection E∩F of the two regions defined by E={(x,y):(x^2 /a^2 )+(y^2 /b^2 )≤1} and F={(x,y):(x^2 /b^2 )+(y^2 /a^2 )≤1} is 4absin^(−1) ((b/(√(a^2 +b^2 )))).$
$S u p p o s e 0 < b ⩽ a . S h o w t h a t t h e a r e a o f$ $i n t e r s e c t i o n E \cap F o f t h e t w o r e g i o n s$ $d e f i n e d b y$ $E = {(x, y) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} ⩽ 1} a n d$ $F = {(x, y) : \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} ⩽ 1}$ $i s 4 {a b s i n}^{- 1} (\frac{b}{\sqrt{a^{2} + b^{2}}}) .$
	Answered by Rasheed Soomro last updated on 08/Nov/15
	$Strategy • E^(⌢) :(x^2 /a^2 )+(y^2 /b^2 )=1 (1) and F^(⌢) :(x^2 /b^2 )+(y^2 /a^2 )=1 (2) are equations of ellipses which are boundary−curves of the regions E and F respectively. •For E∩F≠φ the two ellipses intersect at two points.Let these points are A(x_1 ,y_1 ) and B(x_2 ,y_2 ). •AB^(−) (common chord) divides each of the E and F regions into two parts(segments). • Let e and f are the areas of respective segments of E and F which make E∩F and A =E∩F. Then A=e+f •Let the coordinate system is so changed that A(x_1 ,y_1 ) is origin and x−axis is passed through B in new coordinate system. The coordinates of A and B will be: A=(0,0) and B=(mAB^(−) ,0) mAB^(−) =(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 )) • e is the area between curve and x−axis from x=0 to x=mAB^(−) . So as f. Hence e and f can be determined using definite_(−) integral_(−) of the curve. ∗∗∗∗∗ Determine intersection points A(x_1 ,y_1 ) and B(x_2 ,y_2 ) ⇒y=±((ab)/(√(a^2 +b^2 ))) ⇒x=±((ab)/(√(a^2 +b^2 ))) {(((ab)/(√(a^2 +b^2 ))),((ab)/(√(a^2 +b^2 )))),(((ab)/(√(a^2 +b^2 ))),((−ab)/(√(a^2 +b^2 )))), ( ((−ab)/(√(a^2 +b^2 ))),((ab)/(√(a^2 +b^2 )))),(((−ab)/(√(a^2 +b^2 ))),((−ab)/(√(a^2 +b^2 ))))} −−−−−−−− mAB^(−) =((2ab)/(√(a^2 +b^2 ))) , ((2(√(2 ))ab)/(√(a^2 +b^2 ))) −−−−−−−− e=∫_0 ^(mAB^(−) ) (±b(√(1−(x^2 /a^2 ))) )dx=±(b/a)∫_0 ^(mAB^(−) ) (√(a^2 −x^2 )) dx =±(b/a)∣(x/2)(√(a^2 −x^2 ))+(a^2 /2)sin^(−1) (x/a) +C∣_0 ^(mAB^(−) ) =±(b/a){[((((2ab)/(√(a^2 +b^2 ))) )/2)(√(a^2 −(((2ab)/(√(a^2 +b^2 ))))^2 ))+(a^2 /2)sin^(−1) (((((2ab)/(√(a^2 +b^2 )))))/a) +C] −[(0/2)(√(a^2 −(0)^2 ))+(a^2 /2)sin^(−1) (0/a) +C]} f=∫_0 ^(mAB^(−) ) (±a(√(1−(x^2 /b^2 ))) )dx=±(a/b)∫_0 ^(mAB^(−) ) (√(b^2 −x^2 )) dx Continue$
	$S t r a t e g y$ $∙ \overset{⌢}{E} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (1) a n d \overset{⌢}{F} : \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 (2) a r e e q u a t i o n s o f e l l i p s e s$ $w h i c h a r e b o u n d a r y - c u r v e s o f t h e r e g i o n s E a n d F r e s p e c t i v e l y .$ $∙ F o r E \cap F \neq ϕ t h e t w o e l l i p s e s i n t e r s e c t a t t w o p o i n t s . L e t t h e s e$ $p o i n t s a r e A (x_{1}, y_{1}) a n d B (x_{2}, y_{2}) .$ $∙ \overset{―}{A B} (c o m m o n c h o r d) d i v i d e s e a c h o f t h e E a n d F r e g i o n s$ $i n t o t w o p a r t s (s e g m e n t s) .$ $∙ L e t e a n d f a r e t h e a r e a s o f r e s p e c t i v e s e g m e n t s o f E a n d F w h i c h$ $m a k e E \cap F a n d A = E \cap F . T h e n$ $A = e + f$ $∙ L e t t h e c o o r d i n a t e s y s t e m i s s o c h a n g e d t h a t A (x_{1}, y_{1}) i s$ $o r i g i n a n d x - a x i s i s p a s s e d t h r o u g h B i n n e w c o o r d i n a t e$ $s y s t e m . T h e c o o r d i n a t e s o f A a n d B w i l l b e :$ $A = (0, 0)$ $a n d B = (m \overset{―}{A B}, 0)$ $m \overset{―}{A B} = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ $∙ e i s t h e a r e a b e t w e e n c u r v e a n d x - a x i s f r o m x = 0 t o$ $x = m \overset{―}{A B} .$ $S o a s f .$ $H e n c e e a n d f c a n b e d e t e r m i n e d u s i n g \underline{d e f i n i t e}$ $\underline{i n t e g r a l} o f t h e c u r v e .$ $* * * * *$ $D e t e r m i n e i n t e r s e c t i o n p o i n t s A (x_{1}, y_{1}) a n d B (x_{2}, y_{2})$ $\Rightarrow y = \pm \frac{a b}{\sqrt{a^{2} + b^{2}}}$ $\Rightarrow x = \pm \frac{a b}{\sqrt{a^{2} + b^{2}}}$ ${(\frac{a b}{\sqrt{a^{2} + b^{2}}}, \frac{a b}{\sqrt{a^{2} + b^{2}}}), (\frac{a b}{\sqrt{a^{2} + b^{2}}}, \frac{- a b}{\sqrt{a^{2} + b^{2}}}),$ $(\frac{- a b}{\sqrt{a^{2} + b^{2}}}, \frac{a b}{\sqrt{a^{2} + b^{2}}}), (\frac{- a b}{\sqrt{a^{2} + b^{2}}}, \frac{- a b}{\sqrt{a^{2} + b^{2}}})}$ $- - - - - - - -$ $m \overset{―}{A B} = \frac{2 a b}{\sqrt{a^{2} + b^{2}}}, \frac{2 \sqrt{2} a b}{\sqrt{a^{2} + b^{2}}}$ $- - - - - - - -$ $e = \int_{0}^{m \overset{―}{A B}} (\pm b \sqrt{1 - \frac{x^{2}}{a^{2}}}) d x = \pm \frac{b}{a} \int_{0}^{m \overset{―}{A B}} \sqrt{a^{2} - x^{2}} d x$ $= \pm \frac{b}{a} ∣ \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} \frac{x}{a} + C ∣_{0}^{m \overset{―}{A B}}$ $= \pm \frac{b}{a} {[\frac{\frac{2 a b}{\sqrt{a^{2} + b^{2}}}}{2} \sqrt{a^{2} - {(\frac{2 a b}{\sqrt{a^{2} + b^{2}}})}^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} \frac{(\frac{2 a b}{\sqrt{a^{2} + b^{2}}})}{a} + C]$ $- [\frac{0}{2} \sqrt{a^{2} - {(0)}^{2}} + \frac{a^{2}}{2} {s i n}^{- 1} \frac{0}{a} + C]}$ $f = \int_{0}^{m \overset{―}{A B}} (\pm a \sqrt{1 - \frac{x^{2}}{b^{2}}}) d x = \pm \frac{a}{b} \int_{0}^{m \overset{―}{A B}} \sqrt{b^{2} - x^{2}} d x$ $C o n t i n u e$
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