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Question Number 27643 by ajfour last updated on 11/Jan/18

Commented by Rasheed.Sindhi last updated on 12/Jan/18

$Sir Ajfour, please help me in$ $You can't use 'macro parameter character #' in math mode$
	Answered by mrW2 last updated on 12/Jan/18

	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ${l e t}^{'} s s a y b > a$ $r^{2} (\frac{\cos^{2} θ}{a^{2}} + \frac{\sin^{2} θ}{b^{2}}) = 1$ $r^{2} = \frac{a^{2} b^{2}}{{(a \sin θ)}^{2} + {(b \cos θ)}^{2}} = \frac{a^{2} b^{2}}{b^{2} - (b^{2} - a^{2}) \sin^{2} θ} = \frac{b^{2}}{b^{2} - a^{2}} \times \frac{a^{2}}{\frac{b^{2}}{b^{2} - a^{2}} - \sin^{2} θ} = \frac{c^{2} a^{2}}{c^{2} - \sin^{2} θ}$ $w i t h c = \frac{b}{\sqrt{b^{2} - a^{2}}} > 1$ $\frac{A}{8} = \int d A = \int_{0}^{\frac{π}{4}} \frac{r^{2} d θ}{2} = \frac{c^{2} a^{2}}{2} \int_{0}^{\frac{π}{4}} \frac{d θ}{c^{2} - \sin^{2} θ}$ $= \frac{{c a}^{2}}{2 \sqrt{c^{2} - 1}} {[\tan^{- 1} (\frac{\sqrt{c^{2} - 1}}{c} \tan θ)]}_{0}^{\frac{π}{4}}$ $= \frac{{c a}^{2}}{2 \sqrt{c^{2} - 1}} \tan^{- 1} \frac{\sqrt{c^{2} - 1}}{c}$ $\Rightarrow A = \frac{4 {c a}^{2}}{\sqrt{c^{2} - 1}} \tan^{- 1} \frac{\sqrt{c^{2} - 1}}{c}$ $\sqrt{c^{2} - 1} = \sqrt{\frac{b^{2}}{b^{2} - a^{2}} - 1} = \frac{a}{\sqrt{b^{2} - a^{2}}}$ $\frac{\sqrt{c^{2} - 1}}{c} = \frac{a}{b}$ $\Rightarrow A = 4 a b \tan^{- 1} \frac{a}{b}$
	Commented by ajfour last updated on 12/Jan/18

	$t h a n k y o u s i r, c o r r e c t a n s w e r,$ $a p p r e c i a t e y o u r m e t h o d s i r!$
	Commented by mrW2 last updated on 12/Jan/18

	Commented by ajfour last updated on 12/Jan/18

	$u n d e r s t o o d s i r .$
	Answered by ajfour last updated on 12/Jan/18
	$(c^2 /a^2 )+(c^2 /b^2 )=1 ⇒ c=((ab)/(√(a^2 +b^2 ))) (x^2 /a^2 )+(y^2 /b^2 )=1 ⇒ y=(b/a)(√(a^2 −x^2 )) (A/8)=∫_0 ^( c) [(b/a)(√(a^2 −x^2 )) −x]dx ={(b/a)[(x/2)(√(a^2 −x^2 ))+(a^2 /2)sin^(−1) (x/a)]−(x^2 /2)}∣_0 ^c =(b/a)×(c/2)(√(a^2 −c^2 ))+((ab)/2)sin^(−1) (c/a)−(c^2 /2) =(b/a)×((ab)/(2(√(a^2 +b^2 ))))×(a^2 /(√(a^2 +b^2 )))+ ((ab)/2)sin^(−1) (b/(√(a^2 +b^2 )))−((a^2 b^2 )/(a^2 +b^2 )) (A/8)=((ab)/2)tan^(−1) (b/a) ⇒ A=4abtan^(−1) (b/a) .$
	$\frac{c^{2}}{a^{2}} + \frac{c^{2}}{b^{2}} = 1 \Rightarrow c = \frac{a b}{\sqrt{a^{2} + b^{2}}}$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Rightarrow y = \frac{b}{a} \sqrt{a^{2} - x^{2}}$ $\frac{A}{8} = \int_{0}^{c} [\frac{b}{a} \sqrt{a^{2} - x^{2}} - x] d x$ $= {\frac{b}{a} [\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a}] - \frac{x^{2}}{2}} ∣_{0}^{c}$ $= \frac{b}{a} \times \frac{c}{2} \sqrt{a^{2} - c^{2}} + \frac{a b}{2} \sin^{- 1} \frac{c}{a} - \frac{c^{2}}{2}$ $= \frac{b}{a} \times \frac{a b}{2 \sqrt{a^{2} + b^{2}}} \times \frac{a^{2}}{\sqrt{a^{2} + b^{2}}} +$ $\frac{a b}{2} \sin^{- 1} \frac{b}{\sqrt{a^{2} + b^{2}}} - \frac{a^{2} b^{2}}{a^{2} + b^{2}}$ $\frac{A}{8} = \frac{a b}{2} \tan^{- 1} \frac{b}{a}$ $\Rightarrow A = 4 a b \tan^{- 1} \frac{b}{a} .$
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