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Question Number 41488 by behi83417@gmail.com last updated on 08/Aug/18
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18
(x^2 /a^2 )+(y^2 /(((a^2 /b^2 ))))=1 (x^2 /(((a^2 /b^2 ))))+(y^2 /a^2 )=1 a^2 >(a^2 /b^2 ) (assumed) solve x^2 +b^2 y^2 =a^(2 ) putting the value ofx^2 in2nd eqn b^2 x^2 +y^2 =a^2 b^2 (a^2 −b^2 y^2 )+y^2 =a^2 a^2 b^2 +y^2 (1−b^4 )=a^2 y^2 (1−b^4 )=a^2 (1−b^2 ) y^2 =(a^2 /(1+b^2 )) so y=±(a/( (√(1+b^2 )))) x^2 +b^2 y^2 =a^2 x^2 +b^2 ((a^2 /(1+b^2 )))=a^2 x^2 =a^2 −((a^2 b^2 )/(1+b^2 )) x^2 =((a^2 +a^2 b^2 −a^2 b^2 )/(1+b^2 )) x=±(a/( (√(1+b^2 )))) and y=(a/( (√(1+b^2 )))) contd
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}=\mathrm{1}\:\:\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${a}^{\mathrm{2}} >\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\:\left({assumed}\right) \\ $$$${solve} \\ $$$${x}^{\mathrm{2}} +{b}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}\:} \:\:{putting}\:{the}\:{value}\:{ofx}^{\mathrm{2}} \:{in}\mathrm{2}{nd}\:{eqn} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} {y}^{\mathrm{2}} \right)+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} +{y}^{\mathrm{2}} \left(\mathrm{1}−{b}^{\mathrm{4}} \right)={a}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} \left(\mathrm{1}−{b}^{\mathrm{4}} \right)={a}^{\mathrm{2}} \left(\mathrm{1}−{b}^{\mathrm{2}} \right) \\ $$$${y}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }\:\:\:\:{so}\:{y}=\pm\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }} \\ $$$${x}^{\mathrm{2}} +{b}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }\right)={a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} } \\ $$$${x}=\pm\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\:\:{and}\:{y}=\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }} \\ $$$${contd} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18
=4[∫_0 ^(a/( (√(1+b^2 )))) (1/(b.))(√(a^2 −x^2 )) dx+∫_(a/( (√(1+b^2 )))) ^(a/b) b.(√((a^2 /b^2 )−x^2 )) dx] =4[(1/b)∣(x/2)(√(a^2 −x^2 )) +(a^2 /2)sin^(−1) ((x/a))∣_0 ^(a/( (√(1+b^2 )))) ]+b[∣(x/2)(√((a^2 /b^2 )−x^2 )) +((a^2 /b^2 )/2)sin^(−1) ((x/(a/b)))∣_(a/( (√(1+b^2 )))) ^(a/b) ] =4[(1/b){(a/(2(√(1+b^2 )))).(√(a^2 −(a^2 /(1+b^2 )))) +(a^2 /2)sin^(−1) ((1/( (√(1+b^2 )))))}+ b. [(0+(a^2 /(2b^2 )).(Π/2))−((a/(2(√(1+b^2 ))))(√((a^2 /b^2 )−_ (a^2 /(1+b^2 )))) +(a^2 /(2b^2 ))sin^(−1) (b/( (√(1+b^2 )))))
$$=\mathrm{4}\left[\int_{\mathrm{0}} ^{\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}} \frac{\mathrm{1}}{{b}.}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}+\int_{\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}} ^{\frac{{a}}{{b}}} {b}.\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−{x}^{\mathrm{2}} }\:{dx}\right] \\ $$$$=\mathrm{4}\left[\frac{\mathrm{1}}{{b}}\mid\frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)\overset{\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}} {\mid}_{\mathrm{0}} \right]+{b}\left[\mid\frac{{x}}{\mathrm{2}}\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−{x}^{\mathrm{2}} }\:+\frac{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{\frac{{a}}{{b}}}\right)\overset{\frac{{a}}{{b}}} {\mid}_{\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}} \right] \\ $$$$=\mathrm{4}\left[\frac{\mathrm{1}}{{b}}\left\{\frac{{a}}{\mathrm{2}\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}.\sqrt{{a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\right)\right\}+\right. \\ $$$${b}.\:\left[\left(\mathrm{0}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}{b}^{\mathrm{2}} }.\frac{\Pi}{\mathrm{2}}\right)−\left(\frac{{a}}{\mathrm{2}\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\sqrt{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−_{} \frac{{a}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}{b}^{\mathrm{2}} }{sin}^{−\mathrm{1}} \frac{{b}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\right)\right. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Aug/18
i shall complete the problem in copy then post.. pld wait...
$${i}\:{shall}\:{complete}\:{the}\:{problem}\:{in}\:{copy}\:{then}\:{post}.. \\ $$$${pld}\:{wait}… \\ $$
Commented by behi83417@gmail.com last updated on 08/Aug/18
dear tanmay sir!thank you for so hard work.
$${dear}\:{tanmay}\:{sir}!{thank}\:{you}\:{for} \\ $$$${so}\:{hard}\:{work}. \\ $$
Answered by MJS last updated on 08/Aug/18
(1) x^2 +b^2 y^2 −a^2 =0 (2) b^2 x^2 +y^2 −a^2 =0 ⇒ y_1 =(√(a^2 −b^2 x^2 )) (1−2) (1−b^2 )x^2 +(b^2 −1)y^2 =0 ⇒ x^2 =y^2 in (1) y^2 +b^2 y−a^2 =0 ⇒ y=±(a/( (√(1+b^2 )))) ⇒ x=±(a/( (√(1+b^2 )))) area = square+4×bow square = (((2a)/( (√(1+b^2 )))))^2 =((4a^2 )/(1+b^2 )) 4×bow=2×4×∫_(a/( (√(1+b^2 )))) ^(a/b) y_1 dx=2π(a^2 /b)−((4a^2 )/(1+b^2 ))−((4a^2 )/b)arctan b area = ((2a^2 )/b)(π−2arctan b) (will post the work to ∫y_1 dx later)
$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{b}^{\mathrm{2}} {y}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{y}_{\mathrm{1}} =\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}−\mathrm{2}\right)\:\:\:\:\:\left(\mathrm{1}−{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} +\left({b}^{\mathrm{2}} −\mathrm{1}\right){y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \\ $$$$\mathrm{in}\:\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} +{b}^{\mathrm{2}} {y}−{a}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{y}=\pm\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\:\Rightarrow\:{x}=\pm\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }} \\ $$$$\mathrm{area}\:=\:\mathrm{square}+\mathrm{4}×\mathrm{bow} \\ $$$$\mathrm{square}\:=\:\left(\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}\right)^{\mathrm{2}} =\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} } \\ $$$$\mathrm{4}×\mathrm{bow}=\mathrm{2}×\mathrm{4}×\underset{\frac{{a}}{\:\sqrt{\mathrm{1}+{b}^{\mathrm{2}} }}} {\overset{\frac{{a}}{{b}}} {\int}}{y}_{\mathrm{1}} {dx}=\mathrm{2}\pi\frac{{a}^{\mathrm{2}} }{{b}}−\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{1}+{b}^{\mathrm{2}} }−\frac{\mathrm{4}{a}^{\mathrm{2}} }{{b}}\mathrm{arctan}\:{b} \\ $$$$\mathrm{area}\:=\:\frac{\mathrm{2}{a}^{\mathrm{2}} }{{b}}\left(\pi−\mathrm{2arctan}\:{b}\right) \\ $$$$ \\ $$$$\left(\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{work}\:\mathrm{to}\:\int{y}_{\mathrm{1}} {dx}\:\mathrm{later}\right) \\ $$
Commented by behi83417@gmail.com last updated on 08/Aug/18
nice & smart.thanks in advance sir.
$${nice}\:\&\:{smart}.{thanks}\:{in}\:{advance}\:{sir}. \\ $$

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