Question Number 27643 by ajfour last updated on 11/Jan/18
Commented by Rasheed.Sindhi last updated on 12/Jan/18
$$\mathrm{Sir}\:\mathrm{Ajfour},\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{in} \\ $$$$\mathrm{Q}#\mathrm{27627}\:\&\:\mathrm{Q}#\mathrm{27422} \\ $$
Answered by mrW2 last updated on 12/Jan/18
![(x^2 /a^2 )+(y^2 /b^2 )=1 let′s say b>a r^2 (((cos^2 θ)/a^2 )+((sin^2 θ)/b^2 ))=1 r^2 =((a^2 b^2 )/((a sin θ)^2 +(b cos θ)^2 ))=((a^2 b^2 )/(b^2 −(b^2 −a^2 )sin^2 θ))=(b^2 /(b^2 −a^2 ))×(a^2 /((b^2 /(b^2 −a^2 ))−sin^2 θ))=((c^2 a^2 )/(c^2 −sin^2 θ)) with c=(b/( (√(b^2 −a^2 ))))>1 (A/8)=∫dA=∫_0 ^(π/4) ((r^2 dθ)/2)=((c^2 a^2 )/2)∫_0 ^(π/4) (dθ/(c^2 −sin^2 θ)) =((ca^2 )/(2(√(c^2 −1))))[tan^(−1) (((√(c^2 −1))/c)tan θ)]_0 ^(π/4) =((ca^2 )/(2(√(c^2 −1))))tan^(−1) ((√(c^2 −1))/c) ⇒A=((4ca^2 )/( (√(c^2 −1))))tan^(−1) ((√(c^2 −1))/c) (√(c^2 −1))=(√((b^2 /(b^2 −a^2 ))−1))=(a/( (√(b^2 −a^2 )))) ((√(c^2 −1))/c)=(a/b) ⇒A=4ab tan^(−1) (a/b)](https://www.tinkutara.com/question/Q27672.png)
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${let}'{s}\:{say}\:{b}>{a} \\ $$$${r}^{\mathrm{2}} \left(\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{{b}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${r}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\left({a}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left({b}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{b}^{\mathrm{2}} −\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\mathrm{sin}^{\mathrm{2}} \:\theta}=\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }×\frac{{a}^{\mathrm{2}} }{\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\mathrm{sin}^{\mathrm{2}} \:\theta}=\frac{{c}^{\mathrm{2}} {a}^{\mathrm{2}} }{{c}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$${with}\:{c}=\frac{{b}}{\:\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}>\mathrm{1} \\ $$$$\frac{{A}}{\mathrm{8}}=\int{dA}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{r}^{\mathrm{2}} {d}\theta}{\mathrm{2}}=\frac{{c}^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{d}\theta}{{c}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$=\frac{{ca}^{\mathrm{2}} }{\mathrm{2}\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}{{c}}\mathrm{tan}\:\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{{ca}^{\mathrm{2}} }{\mathrm{2}\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}{{c}} \\ $$$$ \\ $$$$\Rightarrow{A}=\frac{\mathrm{4}{ca}^{\mathrm{2}} }{\:\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}{{c}} \\ $$$$\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}=\sqrt{\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\mathrm{1}}=\frac{{a}}{\:\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$$$\frac{\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}}{{c}}=\frac{{a}}{{b}} \\ $$$$\Rightarrow{A}=\mathrm{4}{ab}\:\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{b}} \\ $$
Commented by ajfour last updated on 12/Jan/18
$${thank}\:{you}\:{sir},\:{correct}\:{answer}, \\ $$$$\:{appreciate}\:{your}\:{method}\:{sir}! \\ $$
Commented by mrW2 last updated on 12/Jan/18
Commented by ajfour last updated on 12/Jan/18
$${understood}\:{sir}\:. \\ $$
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) .](https://www.tinkutara.com/question/Q27679.png)
$$\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\:\:\Rightarrow\:{c}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\Rightarrow\:{y}=\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\frac{{A}}{\mathrm{8}}=\int_{\mathrm{0}} ^{\:\:{c}} \left[\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:−{x}\right]{dx} \\ $$$$=\left\{\frac{{b}}{{a}}\left[\frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{{a}}\right]−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right\}\mid_{\mathrm{0}} ^{{c}} \\ $$$$=\frac{{b}}{{a}}×\frac{{c}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }+\frac{{ab}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{{c}}{{a}}−\frac{{c}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{b}}{{a}}×\frac{{ab}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}×\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+ \\ $$$$\:\:\:\:\:\:\:\:\frac{{ab}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}−\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\frac{{A}}{\mathrm{8}}=\frac{{ab}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}} \\ $$$$\Rightarrow\:\:{A}=\mathrm{4}{ab}\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\:. \\ $$