Question Number 2004 by Yozzi last updated on 29/Oct/15
$${Suppose}\:\mathrm{0}<{b}\leqslant{a}.\:{Show}\:{that}\:{the}\:{area}\:{of} \\ $$$${intersection}\:{E}\cap{F}\:{of}\:{the}\:{two}\:{regions} \\ $$$${defined}\:{by}\: \\ $$$${E}=\left\{\left({x},{y}\right):\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\leqslant\mathrm{1}\right\}\:{and} \\ $$$${F}=\left\{\left({x},{y}\right):\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\leqslant\mathrm{1}\right\}\: \\ $$$${is}\:\:\:\:\mathrm{4}{absin}^{−\mathrm{1}} \left(\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right). \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 08/Nov/15
![Strategy • E^(⌢) :(x^2 /a^2 )+(y^2 /b^2 )=1 (1) and F^(⌢) :(x^2 /b^2 )+(y^2 /a^2 )=1 (2) are equations of ellipses which are boundary−curves of the regions E and F respectively. •For E∩F≠φ the two ellipses intersect at two points.Let these points are A(x_1 ,y_1 ) and B(x_2 ,y_2 ). •AB^(−) (common chord) divides each of the E and F regions into two parts(segments). • Let e and f are the areas of respective segments of E and F which make E∩F and A =E∩F. Then A=e+f •Let the coordinate system is so changed that A(x_1 ,y_1 ) is origin and x−axis is passed through B in new coordinate system. The coordinates of A and B will be: A=(0,0) and B=(mAB^(−) ,0) mAB^(−) =(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 )) • e is the area between curve and x−axis from x=0 to x=mAB^(−) . So as f. Hence e and f can be determined using definite_(−) integral_(−) of the curve. ∗∗∗∗∗ Determine intersection points A(x_1 ,y_1 ) and B(x_2 ,y_2 ) ⇒y=±((ab)/( (√(a^2 +b^2 )))) ⇒x=±((ab)/( (√(a^2 +b^2 )))) {(((ab)/( (√(a^2 +b^2 )))),((ab)/( (√(a^2 +b^2 ))))),(((ab)/( (√(a^2 +b^2 )))),((−ab)/( (√(a^2 +b^2 ))))), ( ((−ab)/( (√(a^2 +b^2 )))),((ab)/( (√(a^2 +b^2 ))))),(((−ab)/( (√(a^2 +b^2 )))),((−ab)/( (√(a^2 +b^2 )))))} −−−−−−−− mAB^(−) =((2ab)/( (√(a^2 +b^2 )))) , ((2(√(2 ))ab)/( (√(a^2 +b^2 )))) −−−−−−−− e=∫_0 ^(mAB^(−) ) (±b(√(1−(x^2 /a^2 ))) )dx=±(b/a)∫_0 ^(mAB^(−) ) (√(a^2 −x^2 )) dx =±(b/a)∣(x/2)(√(a^2 −x^2 ))+(a^2 /2)sin^(−1) (x/a) +C∣_0 ^(mAB^(−) ) =±(b/a){[((((2ab)/( (√(a^2 +b^2 )))) )/2)(√(a^2 −(((2ab)/( (√(a^2 +b^2 )))))^2 ))+(a^2 /2)sin^(−1) (((((2ab)/( (√(a^2 +b^2 ))))))/a) +C] −[(0/2)(√(a^2 −(0)^2 ))+(a^2 /2)sin^(−1) (0/a) +C]} f=∫_0 ^(mAB^(−) ) (±a(√(1−(x^2 /b^2 ))) )dx=±(a/b)∫_0 ^(mAB^(−) ) (√(b^2 −x^2 )) dx Continue](https://www.tinkutara.com/question/Q2162.png)
$${Strategy} \\ $$$$\bullet\:\overset{\frown} {{E}}:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\left(\mathrm{1}\right)\:\:{and}\:\:\:\overset{\frown} {{F}}:\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1}\:\left(\mathrm{2}\right)\:{are}\:{equations}\:{of}\:{ellipses}\:\: \\ $$$${which}\:{are}\:{boundary}−{curves}\:{of}\:{the}\:{regions}\:{E}\:{and}\:{F}\:{respectively}. \\ $$$$\bullet{For}\:\:{E}\cap{F}\neq\phi\:{the}\:{two}\:{ellipses}\:{intersect}\:{at}\:{two}\:{points}.{Let}\:{these}\: \\ $$$${points}\:{are}\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:{and}\:\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right). \\ $$$$\bullet\overline {{AB}}\:\left({common}\:{chord}\right)\:{divides}\:{each}\:{of}\:{the}\:{E}\:{and}\:{F}\:{regions}\: \\ $$$${into}\:{two}\:{parts}\left({segments}\right). \\ $$$$\bullet\:{Let}\:{e}\:{and}\:{f}\:{are}\:{the}\:{areas}\:{of}\:\:{respective}\:{segments}\:{of}\:{E}\:{and}\:{F}\:{which} \\ $$$${make}\:{E}\cap{F}\:{and}\:\boldsymbol{\mathrm{A}}\:={E}\cap{F}.\:{Then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{A}}={e}+{f} \\ $$$$\bullet{Let}\:\:{the}\:{coordinate}\:{system}\:{is}\:{so}\:{changed}\:{that}\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:{is}\: \\ $$$${origin}\:{and}\:\:\:{x}−{axis}\:{is}\:{passed}\:{through}\:{B}\:{in}\:{new}\:{coordinate}\: \\ $$$${system}.\:{The}\:{coordinates}\:{of}\:{A}\:{and}\:{B}\:{will}\:{be}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}=\left(\mathrm{0},\mathrm{0}\right)\:\:\:\: \\ $$$$\:\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\:\:\:\:\:{B}=\left({m}\overline {{AB}},\mathrm{0}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}\overline {{AB}}=\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$\:\bullet\:{e}\:{is}\:{the}\:{area}\:{between}\:{curve}\:{and}\:{x}−{axis}\:{from}\:{x}=\mathrm{0}\:{to} \\ $$$${x}={m}\overline {{AB}}. \\ $$$${So}\:{as}\:{f}. \\ $$$$\:\:\:\:\:\:\:\:\:{Hence}\:\:{e}\:{and}\:{f}\:\:{can}\:{be}\:{determined}\:{using}\underset{−} {\:{definite}} \\ $$$$\underset{−} {{integral}}\:{of}\:{the}\:{curve}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\ast\ast\ast\ast \\ $$$${Determine}\:{intersection}\:{points}\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:{and}\:\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{y}=\pm\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow{x}=\pm\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\left\{\left(\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right),\left(\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right),\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right),\left(\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)\right\} \\ $$$$−−−−−−−− \\ $$$${m}\overline {{AB}}=\frac{\mathrm{2}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:\:,\:\frac{\mathrm{2}\sqrt{\mathrm{2}\:}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\: \\ $$$$−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:{e}=\int_{\mathrm{0}} ^{{m}\overline {{AB}}} \left(\pm{b}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:\right){dx}=\pm\frac{{b}}{{a}}\int_{\mathrm{0}} ^{{m}\overline {{AB}}} \:\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\pm\frac{{b}}{{a}}\mid\frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \frac{{x}}{{a}}\:+{C}\mid_{\mathrm{0}} ^{{m}\overline {{AB}}} \\ $$$$\:\:\:=\pm\frac{{b}}{{a}}\left\{\left[\frac{\frac{\mathrm{2}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −\left(\frac{\mathrm{2}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \frac{\left(\frac{\mathrm{2}{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right)}{{a}}\:+{C}\right]\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\frac{\mathrm{0}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −\left(\mathrm{0}\right)^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \frac{\mathrm{0}}{{a}}\:+{C}\right]\right\} \\ $$$$\:\:\:\:\:\:\:\:{f}=\int_{\mathrm{0}} ^{{m}\overline {{AB}}} \left(\pm{a}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}\:\right){dx}=\pm\frac{{a}}{{b}}\int_{\mathrm{0}} ^{{m}\overline {{AB}}} \:\sqrt{{b}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}\:\:\: \\ $$$${Continue} \\ $$