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Question Number 177451 by mr W last updated on 05/Oct/22

Commented by mr W last updated on 05/Oct/22

$a n e l i p s e w i t h a = 10 c m a n d b = 6 c m i s$ $m o v e d b y 5 c m i n x d i r e c t i o n a n d b y$ $3 c m i n y d i r e c t i o n . f i n d t h e s h a d e d$ $a r e a .$
	Answered by mr W last updated on 05/Oct/22

	Commented by mr W last updated on 05/Oct/22
	$red elipse: (x^2 /(10^2 ))+(y^2 /6^2 )=1 ⇒y=6(√(1−(x^2 /(100)))) blue elipse: (((x−5)^2 )/(10^2 ))+(((y−3)^2 )/6^2 )=1 line PQ: ((10x−25)/(10^2 ))+((6y−9)/6^2 )=0 ⇒y=3(1−(x/5)) point P and Q: (x^2 /(10^2 ))+(3^2 /6^2 )(1−(x/5))^2 =1 2x^2 −10x−75=0 x=((5±5(√7))/2) ⇒x_Q =((5(1+(√7)))/2), y_Q =((3(1−(√7)))/2) ⇒x_P =((5(1−(√7)))/2), y_P =((3(1+(√7)))/2) A_(segment) =∫_x_P ^x_Q [6(√(1−(x^2 /(100))))−3+((3x)/5)]dx+2∫_x_Q ^(10) 6(√(1−(x^2 /(100)))) dx =3∫_x_P ^x_Q [2(√(1−(x^2 /(100))))−1+(x/5)]dx+12∫_x_Q ^(10) (√(1−(x^2 /(100)))) dx =3[x(√(1−(x^2 /(100))))+10 sin^(−1) (x/(10))−x+(x^2 /(10))]_x_P ^x_Q +6[x(√(1−(x^2 /(100))))+10 sin^(−1) (x/(10))]_x_Q ^(10) =3{((5(1+(√7)))/2)(√(1−(((1+(√7))^2 )/(16))))−((5(1−(√7)))/2)(√(1−(((1−(√7))^2 )/(16))))+10( sin^(−1) ((1+(√7))/4)−sin^(−1) ((1−(√7))/4))−((5(1+(√7)−1+(√7)))/2)+((5[(1+(√7))^2 −(1−(√7))^2 ])/8)}+6{10(√(1−((100)/(100))))−((5(1+(√7)))/2)(√(1−(((1+(√7))^2 )/(16))))+10(sin^(−1) ((10)/(10))−sin^(−1) ((1+(√7))/4))} =((45)/2)+30( sin^(−1) ((1+(√7))/4)−sin^(−1) ((1−(√7))/4))−((15(√7))/2)−((45)/2)+30π−60 sin^(−1) ((1+(√7))/4) =30π−30( sin^(−1) ((1+(√7))/4)+ sin^(−1) ((1−(√7))/4))−((15(√7))/2) A_(elipse) =πab=10×6π=60π A_(shaded) =A_(elipse) −2A_(segment) =60π−60π+60(sin^(−1) ((1+(√7))/4)+sin^(−1) ((1−(√7))/4))+15(√7) =60(sin^(−1) ((1+(√7))/4)+ sin^(−1) ((1−(√7))/4))+15(√7) =60 cos^(−1) (3/4)+15(√7) ✓ ≈83.05$
	$r e d e l i p s e :$ $\frac{x^{2}}{10^{2}} + \frac{y^{2}}{6^{2}} = 1 \Rightarrow y = 6 \sqrt{1 - \frac{x^{2}}{100}}$ $b l u e e l i p s e :$ $\frac{{(x - 5)}^{2}}{10^{2}} + \frac{{(y - 3)}^{2}}{6^{2}} = 1$ $l i n e P Q :$ $\frac{10 x - 25}{10^{2}} + \frac{6 y - 9}{6^{2}} = 0$ $\Rightarrow y = 3 (1 - \frac{x}{5})$ $p o i n t P a n d Q :$ $\frac{x^{2}}{10^{2}} + \frac{3^{2}}{6^{2}} {(1 - \frac{x}{5})}^{2} = 1$ $2 x^{2} - 10 x - 75 = 0$ $x = \frac{5 \pm 5 \sqrt{7}}{2}$ $\Rightarrow x_{Q} = \frac{5 (1 + \sqrt{7})}{2}, y_{Q} = \frac{3 (1 - \sqrt{7})}{2}$ $\Rightarrow x_{P} = \frac{5 (1 - \sqrt{7})}{2}, y_{P} = \frac{3 (1 + \sqrt{7})}{2}$ $A_{s e g m e n t} = \int_{x_{P}}^{x_{Q}} [6 \sqrt{1 - \frac{x^{2}}{100}} - 3 + \frac{3 x}{5}] d x + 2 \int_{x_{Q}}^{10} 6 \sqrt{1 - \frac{x^{2}}{100}} d x$ $= 3 \int_{x_{P}}^{x_{Q}} [2 \sqrt{1 - \frac{x^{2}}{100}} - 1 + \frac{x}{5}] d x + 12 \int_{x_{Q}}^{10} \sqrt{1 - \frac{x^{2}}{100}} d x$ $= 3 {[x \sqrt{1 - \frac{x^{2}}{100}} + 10 \sin^{- 1} \frac{x}{10} - x + \frac{x^{2}}{10}]}_{x_{P}}^{x_{Q}} + 6 {[x \sqrt{1 - \frac{x^{2}}{100}} + 10 \sin^{- 1} \frac{x}{10}]}_{x_{Q}}^{10}$ $= 3 {\frac{5 (1 + \sqrt{7})}{2} \sqrt{1 - \frac{{(1 + \sqrt{7})}^{2}}{16}} - \frac{5 (1 - \sqrt{7})}{2} \sqrt{1 - \frac{{(1 - \sqrt{7})}^{2}}{16}} + 10 (\sin^{- 1} \frac{1 + \sqrt{7}}{4} - \sin^{- 1} \frac{1 - \sqrt{7}}{4}) - \frac{5 (1 + \sqrt{7} - 1 + \sqrt{7})}{2} + \frac{5 [{(1 + \sqrt{7})}^{2} - {(1 - \sqrt{7})}^{2}]}{8}} + 6 {10 \sqrt{1 - \frac{100}{100}} - \frac{5 (1 + \sqrt{7})}{2} \sqrt{1 - \frac{{(1 + \sqrt{7})}^{2}}{16}} + 10 (\sin^{- 1} \frac{10}{10} - \sin^{- 1} \frac{1 + \sqrt{7}}{4})}$ $= \frac{45}{2} + 30 (\sin^{- 1} \frac{1 + \sqrt{7}}{4} - \sin^{- 1} \frac{1 - \sqrt{7}}{4}) - \frac{15 \sqrt{7}}{2} - \frac{45}{2} + 30 π - 60 \sin^{- 1} \frac{1 + \sqrt{7}}{4}$ $= 30 π - 30 (\sin^{- 1} \frac{1 + \sqrt{7}}{4} + \sin^{- 1} \frac{1 - \sqrt{7}}{4}) - \frac{15 \sqrt{7}}{2}$ $A_{e l i p s e} = π a b = 10 \times 6 π = 60 π$ $A_{s h a d e d} = A_{e l i p s e} - 2 A_{s e g m e n t}$ $= 60 π - 60 π + 60 (\sin^{- 1} \frac{1 + \sqrt{7}}{4} + \sin^{- 1} \frac{1 - \sqrt{7}}{4}) + 15 \sqrt{7}$ $= 60 (\sin^{- 1} \frac{1 + \sqrt{7}}{4} + \sin^{- 1} \frac{1 - \sqrt{7}}{4}) + 15 \sqrt{7}$ $= 60 \cos^{- 1} \frac{3}{4} + 15 \sqrt{7} ✓$ $\approx 83.05$
	Commented by Ar Brandon last updated on 05/Oct/22

	$Sir, why . . . + 2 \int_{x_{Q}}^{10} 6 \sqrt{1 - \frac{x^{2}}{100}} d x at A_{segment} ?$ $Why not just . . . + \int_{x_{Q}}^{10} 6 \sqrt{1 - \frac{x^{2}}{100}} d x$
	Commented by mr W last updated on 05/Oct/22

	$s e e d i a g r a m :$
	Commented by mr W last updated on 05/Oct/22

	Commented by Ar Brandon last updated on 05/Oct/22

	$I see, Sir . But I {don}^{'} t understand why$ $you multiplied it by 2$
	Commented by mr W last updated on 06/Oct/22

	$t h i s i s o b v i o u s l y w r o n g s i r!$ $y o u c a n c o u n t t h e a r e a f r o m f o l l o w i n g$ $g r a p h . e a c h s m a l l s q u a r e h a s a n a r e a$ $o f 4 .$ $w e c o u n t a p p r o x i m a t e l y 21 s m a l l$ $s q u a r e s, s o t h e a r e a s h o u l d b e$ $a p p r o x i m a t e l y \approx 21 \times 4 = 84$
	Commented by mr W last updated on 05/Oct/22

	Commented by mr W last updated on 05/Oct/22

	$\int_{x_{Q}}^{10} y d x i s o n l y t h e a r e a a b o v e t h e$ $x a x i s . b u t w e h a v e a l s o t h e a r e a$ $u n d e r t h e x a x i s, t h e r e f o r e 2 \int_{x_{Q}}^{10} y d x .$
	Commented by Ar Brandon last updated on 05/Oct/22
	OK got it now. Thanks!
	Commented by a.lgnaoui last updated on 05/Oct/22

	$y_{1}^{2} = \frac{6^{2}}{10^{2}} (100 - x^{2}) (1)$ ${(y_{2} + 3)}^{2} = \frac{6^{2}}{10^{2}} [(100 - {(x + 5)}^{2}] (2)$ $y_{1} = y_{2} (p o i n t s c o m m u n s (1) e t (2)$ $(1) y_{1} = \frac{6}{10} \sqrt{100 - x^{2}}$ ${(y + 3)}^{2} = {[3 + \frac{3}{10} \sqrt{100 - x^{2}}]}^{2} = \frac{9}{25} [(100 - {(x + 5)}^{2}]$ $(9 + (\frac{9}{25}) (100 - x^{2}) + \frac{18}{5} \sqrt{100 - x^{2}}) = \frac{9}{25} (75 - x^{2} - 10 x)$ $(45 - \frac{9 x^{2}}{25} + \frac{18}{5} \sqrt{100 - x^{2}}) = (27 - \frac{9 x^{2}}{25} - \frac{18}{5})$ $1 + \frac{1}{5} \sqrt{100 - x^{2}} + \frac{1}{5} = 0$ $x = 9, 92 \Rightarrow i n t e r s e c t i o n = (4, 92; 9, 92)$ $y = \frac{3}{5} \sqrt{100 - 24, 20} = 5, 22$ $P (4, 92; 5, 22) Q (9, 92; - 2, 25)$ $A u r e = \int \frac{6}{5} + \frac{1}{5} \sqrt{100 - x^{2}} d x$ $= \frac{6}{5} x + 2 \int \sqrt{1 - (\frac{x^{2}}{10})} d x$ $\sin t = \frac{x}{10} d x = 10 \cos t d t = 10 \sqrt{1 - t^{2}} d t$ $= 10 \int (1 - t^{2}) d t = 10 t - 10 \frac{t^{3}}{3}$ $A i r e = \int_{0}^{5} = {[\frac{6 x}{5}]}_{0}^{5} + 10 {[\sin^{- 1} (\frac{x}{10})]}_{0}^{5} - \frac{10}{3} {[{(\sin^{- 1} (\frac{x}{10}))}^{3}]}_{0}^{5}$ $= 6 + {10 \sin}^{- 1} (\frac{1}{2}) - \frac{10}{3} {[\sin^{- 1} (\frac{1}{2})]}^{3}$ $6 + 2 [10 \frac{π}{6} - \frac{10}{3} \times \frac{π^{3}}{216}] = 6 + 21, 44 - 0, 86$ $A i r e = 26, 58$ $b j$
	Commented by mr W last updated on 05/Oct/22

	Commented by a.lgnaoui last updated on 05/Oct/22

	$t h a n k y o u v e r y m u c h$
	Commented by Tawa11 last updated on 06/Oct/22

	$Great sir$
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