Suppose-0-lt-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-

Question Number 2004 by Yozzi last updated on 29/Oct/15

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

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

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