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Question Number 30820 by ajfour last updated on 26/Feb/18

	Answered by mrW2 last updated on 26/Feb/18
	$Eqn. of ellipse 1 (original one): (x^2 /a^2 )+(y^2 /b^2 )=1 ((r^2 cos^2 ϕ)/a^2 )+((r^2 sin^2 ϕ)/b^2 )=1 ⇒r^2 [((cos^2 ϕ)/a^2 )+((sin^2 ϕ)/b^2 )]=1 ⇒r^2 =((a^2 b^2 )/(a^2 sin^2 ϕ+b^2 cos^2 ϕ)) Eqn. of ellipse 2 (rotated one): ⇒r^2 [((cos^2 (ϕ−θ))/a^2 )+((sin^2 (ϕ−θ))/b^2 )]=1 ⇒r^2 =((a^2 b^2 )/(a^2 sin^2 (ϕ−θ)+b^2 cos^2 (ϕ−θ))) Intersection of both ellipses: ((cos^2 ϕ)/a^2 )+((sin^2 ϕ)/b^2 )=((cos^2 (ϕ−θ))/a^2 )+((sin^2 (ϕ−θ))/b^2 ) b^2 cos^2 ϕ+a^2 sin^2 ϕ=b^2 cos^2 (ϕ−θ)+a^2 sin^2 (ϕ−θ) b^2 cos^2 ϕ+a^2 sin^2 ϕ=b^2 (cos ϕ cos θ+sin ϕ sin θ)^2 +a^2 (sin ϕ cos θ−cos ϕ sin θ)^2 b^2 cos^2 ϕ+a^2 sin^2 ϕ=b^2 cos^2 ϕ cos^2 θ+b^2 sin^2 ϕ sin^2 θ+2b^2 sin ϕ cos ϕ sin θ cos θ+a^2 sin^2 ϕ cos^2 θ+a^2 cos^2 ϕ sin^2 θ−2a^2 sin ϕ cos ϕ sin θ cos θ b^2 (cos^2 ϕ−sin^2 ϕ) sin^2 θ−a^2 (cos^2 ϕ−sin^2 ϕ) sin^2 θ=2(b^2 −a^2 ) sin ϕ cos ϕ sin θ cos θ (b^2 −a^2 )cos 2ϕ sin^2 θ=(b^2 −a^2 ) sin 2ϕ sin θ cos θ cos 2ϕ sin θ= sin 2ϕ cos θ ⇒tan 2ϕ=tan θ ⇒2ϕ=θ or θ+π ⇒ϕ_1 =(θ/2) ⇒ϕ_2 =(θ/2)+(π/2) A=∫_ϕ_1 ^( ϕ_2 ) ∫_r_1 ^r_2 rdrdϕ =(1/2)∫_ϕ_1 ^( ϕ_2 ) [r_2 ^2 −r_1 ^2 ]dϕ =((a^2 b^2 )/2)∫_ϕ_1 ^( ϕ_2 ) [(1/(a^2 sin^2 (ϕ−θ)+b^2 cos^2 (ϕ−θ)))−(1/(a^2 sin^2 ϕ+b^2 cos^2 ϕ))]dϕ =((a^2 b^2 )/2)[(1/(ab))tan^(−1) {(a/b) tan (ϕ−θ)}−(1/(ab))tan^(−1) {(a/b) tan ϕ}]_ϕ_1 ^ϕ_2 =((ab)/2)[tan^(−1) {(a/b) tan (ϕ−θ)}−tan^(−1) {(a/b) tan ϕ}]_ϕ_1 ^ϕ_2 =((ab)/2)[tan^(−1) {(a/b) tan ((π/2)−(θ/2))}−tan^(−1) {(a/b) tan ((π/2)+(θ/2))}−tan^(−1) {(a/b) tan ((θ/2)−θ)}+tan^(−1) {(a/b) tan (θ/2)}] =((ab)/2)[2 tan^(−1) {(a/b) tan ((π/2)−(θ/2))}−π+2 tan^(−1) {(a/b) tan (θ/2)}] =ab[tan^(−1) {(a/b) tan ((π/2)−(θ/2))}+tan^(−1) {(a/b) tan (θ/2)}−(π/2)] =ab{(π/2)− tan^(−1) [((2ab)/((a^2 −b^2 )sin θ))]} ⇒A=ab tan^(−1) ∣(((a^2 −b^2 )sin θ)/(2ab))∣ An other way: A=(1/2)(πab−4∫_ϕ_1 ^( ϕ_2 ) (r_1 ^2 /2)dϕ) =ab(π/2)−∫_ϕ_1 ^( ϕ_2 ) ((a^2 b^2 )/(a^2 sin^2 ϕ+b^2 cos^2 ϕ))dϕ =ab(π/2)−a^2 b^2 [(1/(ab)) tan^(−1) {(a/b)tan ϕ}]_ϕ_1 ^ϕ_2 =ab(π/2)−ab[tan^(−1) {(a/b)tan ((π/2)+(θ/2))}−tan^(−1) {(a/b)tan ((θ/2))}] =ab(π/2)−ab[π−tan^(−1) {(a/b)tan ((π/2)−(θ/2))}−tan^(−1) {(a/b)tan ((θ/2))}] =ab[tan^(−1) {(a/b)tan ((π/2)−(θ/2))}+tan^(−1) {(a/b)tan ((θ/2))}−(π/2)] ..... as above$
	$E q n . o f e l l i p s e 1 (o r i g i n a l o n e) :$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{r^{2} \cos^{2} φ}{a^{2}} + \frac{r^{2} \sin^{2} φ}{b^{2}} = 1$ $\Rightarrow r^{2} [\frac{\cos^{2} φ}{a^{2}} + \frac{\sin^{2} φ}{b^{2}}] = 1$ $\Rightarrow r^{2} = \frac{a^{2} b^{2}}{a^{2} \sin^{2} φ + b^{2} \cos^{2} φ}$ $E q n . o f e l l i p s e 2 (r o t a t e d o n e) :$ $\Rightarrow r^{2} [\frac{\cos^{2} (φ - θ)}{a^{2}} + \frac{\sin^{2} (φ - θ)}{b^{2}}] = 1$ $\Rightarrow r^{2} = \frac{a^{2} b^{2}}{a^{2} \sin^{2} (φ - θ) + b^{2} \cos^{2} (φ - θ)}$ $I n t e r s e c t i o n o f b o t h e l l i p s e s :$ $\frac{\cos^{2} φ}{a^{2}} + \frac{\sin^{2} φ}{b^{2}} = \frac{\cos^{2} (φ - θ)}{a^{2}} + \frac{\sin^{2} (φ - θ)}{b^{2}}$ $b^{2} \cos^{2} φ + a^{2} \sin^{2} φ = b^{2} \cos^{2} (φ - θ) + a^{2} \sin^{2} (φ - θ)$ $b^{2} \cos^{2} φ + a^{2} \sin^{2} φ = b^{2} {(\cos φ \cos θ + \sin φ \sin θ)}^{2} + a^{2} {(\sin φ \cos θ - \cos φ \sin θ)}^{2}$ $b^{2} \cos^{2} φ + a^{2} \sin^{2} φ = b^{2} \cos^{2} φ \cos^{2} θ + b^{2} \sin^{2} φ \sin^{2} θ + 2 b^{2} \sin φ \cos φ \sin θ \cos θ + a^{2} \sin^{2} φ \cos^{2} θ + a^{2} \cos^{2} φ \sin^{2} θ - 2 a^{2} \sin φ \cos φ \sin θ \cos θ$ $b^{2} (\cos^{2} φ - \sin^{2} φ) \sin^{2} θ - a^{2} (\cos^{2} φ - \sin^{2} φ) \sin^{2} θ = 2 (b^{2} - a^{2}) \sin φ \cos φ \sin θ \cos θ$ $(b^{2} - a^{2}) \cos 2 φ \sin^{2} θ = (b^{2} - a^{2}) \sin 2 φ \sin θ \cos θ$ $\cos 2 φ \sin θ = \sin 2 φ \cos θ$ $\Rightarrow \tan 2 φ = \tan θ$ $\Rightarrow 2 φ = θ o r θ + π$ $\Rightarrow φ_{1} = \frac{θ}{2}$ $\Rightarrow φ_{2} = \frac{θ}{2} + \frac{π}{2}$ $A = \int_{φ_{1}}^{φ_{2}} \int_{r_{1}}^{r_{2}} r d r d φ$ $= \frac{1}{2} \int_{φ_{1}}^{φ_{2}} [r_{2}^{2} - r_{1}^{2}] d φ$ $= \frac{a^{2} b^{2}}{2} \int_{φ_{1}}^{φ_{2}} [\frac{1}{a^{2} \sin^{2} (φ - θ) + b^{2} \cos^{2} (φ - θ)} - \frac{1}{a^{2} \sin^{2} φ + b^{2} \cos^{2} φ}] d φ$ $= \frac{a^{2} b^{2}}{2} {[\frac{1}{a b} \tan^{- 1} {\frac{a}{b} \tan (φ - θ)} - \frac{1}{a b} \tan^{- 1} {\frac{a}{b} \tan φ}]}_{φ_{1}}^{φ_{2}}$ $= \frac{a b}{2} {[\tan^{- 1} {\frac{a}{b} \tan (φ - θ)} - \tan^{- 1} {\frac{a}{b} \tan φ}]}_{φ_{1}}^{φ_{2}}$ $= \frac{a b}{2} [\tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} - \frac{θ}{2})} - \tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} + \frac{θ}{2})} - \tan^{- 1} {\frac{a}{b} \tan (\frac{θ}{2} - θ)} + \tan^{- 1} {\frac{a}{b} \tan \frac{θ}{2}}]$ $= \frac{a b}{2} [2 \tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} - \frac{θ}{2})} - π + 2 \tan^{- 1} {\frac{a}{b} \tan \frac{θ}{2}}]$ $= a b [\tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} - \frac{θ}{2})} + \tan^{- 1} {\frac{a}{b} \tan \frac{θ}{2}} - \frac{π}{2}]$ $= a b {\frac{π}{2} - \tan^{- 1} [\frac{2 a b}{(a^{2} - b^{2}) \sin θ}]}$ $\Rightarrow A = a b \tan^{- 1} ∣ \frac{(a^{2} - b^{2}) \sin θ}{2 a b} ∣$ $A n o t h e r w a y :$ $A = \frac{1}{2} (π a b - 4 \int_{φ_{1}}^{φ_{2}} \frac{r_{1}^{2}}{2} d φ)$ $= a b \frac{π}{2} - \int_{φ_{1}}^{φ_{2}} \frac{a^{2} b^{2}}{a^{2} \sin^{2} φ + b^{2} \cos^{2} φ} d φ$ $= a b \frac{π}{2} - a^{2} b^{2} {[\frac{1}{a b} \tan^{- 1} {\frac{a}{b} \tan φ}]}_{φ_{1}}^{φ_{2}}$ $= a b \frac{π}{2} - a b [\tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} + \frac{θ}{2})} - \tan^{- 1} {\frac{a}{b} \tan (\frac{θ}{2})}]$ $= a b \frac{π}{2} - a b [π - \tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} - \frac{θ}{2})} - \tan^{- 1} {\frac{a}{b} \tan (\frac{θ}{2})}]$ $= a b [\tan^{- 1} {\frac{a}{b} \tan (\frac{π}{2} - \frac{θ}{2})} + \tan^{- 1} {\frac{a}{b} \tan (\frac{θ}{2})} - \frac{π}{2}]$ $. . . . . a s a b o v e$
	Commented by ajfour last updated on 26/Feb/18

	$O^{'} MY G o d! Y o u m a k e i t p o s s i b l e$ $S i r . T h a n k y o u i m m e n s e l y .$ $I t s r e a l l y g r e a t, S i r .$
	Commented by ajfour last updated on 26/Feb/18

	$Y e s S i r, t h a n k y o u a g a i n .$
	Commented by mrW2 last updated on 26/Feb/18

	$T h a n k s f o r c h e c k i n g!$ $I c h e c k e d a g a i n . T h e f i n a l f o r m u l a i s$ $A = a b \tan^{- 1} ∣ \frac{(a^{2} - b^{2}) \sin θ}{2 a b} ∣$ ${I t}^{'} s m o r e s i m p l e t h a n I^{'} v e e x p e c t e d .$
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