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Question-47559

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Question Number 47559 by ajfour last updated on 11/Nov/18
Commented by ajfour last updated on 11/Nov/18
Q.47519 (Solution attempted)
$${Q}.\mathrm{47519}\:\:\left({Solution}\:{attempted}\right) \\ $$
Answered by ajfour last updated on 11/Nov/18
Area △ABC = (1/2)(AB)(CP) AB = (√(x_A ^2 +y_B ^2 )) Let ∠OAB=θ ; CP = l P (x_1 ,y_1 ), C(x_C ,y_C ) Eq. of tangent to Ellipse at P ((xx_1 )/a^2 )+((yy_1 )/b^2 ) = 1 tan θ = ((b^2 x_1 )/(a^2 y_1 )) ⇒ (x_1 /y_1 ) = ((a^2 sin θ)/(b^2 cos θ)) ..(i) ⇒ x_A = (a^2 /x_1 ) ; y_B = (b^2 /y_1 ) x_C = x_1 −lsin θ y_C = y_1 −lcos θ And since C lies on ellipse (((x_1 −lsin θ)^2 )/a^2 )+(((y_1 −lcos θ)^2 )/b^2 ) = 1 ⇒ ((l^2 sin^2 θ)/a^2 )+((l^2 cos^2 θ)/b^2 )+((x_1 ^2 /a^2 )+(y_1 ^2 /b^2 )) −2l(((x_1 sin θ)/a^2 )+((y_1 cos θ)/b^2 )) = 1 but ((x_1 ^2 /a^2 )+(y_1 ^2 /b^2 )) = 1, So ⇒ l= ((2(((x_1 sin θ)/a^2 )+((y_1 cos θ)/b^2 )))/(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 ))) △ = (l/2)(√(x_A ^2 +y_A ^2 )) = (((((x_1 sin θ)/a^2 )+((y_1 cos θ)/b^2 ))(√((a^4 /x_1 ^2 )+(b^4 /y_1 ^2 ))))/(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 ))) ⇒ △=(((((x_1 sin θ)/(y_1 a^2 ))+((cos θ)/b^2 ))(√(((y_1 ^2 a^4 )/x_1 ^2 )+b^4 )))/(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 ))) using (i) : (x_1 /y_1 ) = ((a^2 sin θ)/(b^2 cos θ)) △ = (((((sin^2 θ)/(b^2 cos θ))+((cos θ)/b^2 ))(√(((b^4 cos^2 θ)/(sin^2 θ))+b^4 )))/(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 ))) △ = (((1/(b^2 cos θ))×(b^2 /(sin θ)))/(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 ))) ⇒ (1/△) = sin θcos θ(((sin^2 θ)/a^2 )+((cos^2 θ)/b^2 )) let tan θ = m ⇒ (1/△) = ((m((m^2 /a^2 )+(1/b^2 )))/((1+m^2 )^2 )) ((d(1/△))/dθ) = ((d(1/△))/dm)×(dm/dθ) ((d(1/△))/dθ) = 0 (for minimum △) ⇒ ((d(1/△))/dm) = (1+m^2 )^2 (((3m^2 )/a^2 )+(1/b^2 )) − 4m^2 (1+m^2 )((m^2 /a^2 )+(1/b^2 )) = 0 ⇒(1+m^2 )(((3m^2 )/a^2 )+(1/b^2 ))=4m^2 ((m^2 /a^2 )+(1/b^2 )) ((3m^2 )/a^2 )+(1/b^2 )+((3m^4 )/a^2 )+(m^2 /b^2 ) = ((4m^4 )/a^2 )+((4m^2 )/b^2 ) ⇒ m^4 +3m^2 ((a^2 /b^2 )−1)−(a^2 /b^2 ) = 0 m^2 = (√((9/4)((a^2 /b^2 )−1)^2 +(a^2 /b^2 ))) −(3/2)((a^2 /b^2 )−1) △_(min) = ((a^2 (1+m^2 )^2 )/(m(m^2 +(a^2 /b^2 )))) . (As a special case) If a= b =R , then m= 1 △_(min) = 2R^2 .
$${Area}\:\bigtriangleup{ABC}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({AB}\right)\left({CP}\right) \\ $$$${AB}\:=\:\sqrt{{x}_{{A}} ^{\mathrm{2}} +{y}_{{B}} ^{\mathrm{2}} } \\ $$$${Let}\:\:\:\angle{OAB}=\theta\:;\:\:{CP}\:=\:{l}\: \\ $$$$\:{P}\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right),\:\:{C}\left({x}_{{C}} \:,{y}_{{C}} \right) \\ $$$${Eq}.\:{of}\:{tangent}\:{to}\:{Ellipse}\:{at}\:{P} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{xx}_{\mathrm{1}} }{{a}^{\mathrm{2}} }+\frac{{yy}_{\mathrm{1}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\mathrm{tan}\:\theta\:=\:\frac{{b}^{\mathrm{2}} {x}_{\mathrm{1}} }{{a}^{\mathrm{2}} {y}_{\mathrm{1}} }\:\:\:\Rightarrow\:\:\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }\:=\:\frac{{a}^{\mathrm{2}} \mathrm{sin}\:\theta}{{b}^{\mathrm{2}} \mathrm{cos}\:\theta}\:\:..\left({i}\right) \\ $$$$\Rightarrow\:\:{x}_{{A}} \:=\:\frac{{a}^{\mathrm{2}} }{{x}_{\mathrm{1}} }\:\:;\:\:{y}_{{B}} \:=\:\frac{{b}^{\mathrm{2}} }{{y}_{\mathrm{1}} } \\ $$$$\:\:\:\:{x}_{{C}} \:=\:{x}_{\mathrm{1}} −{l}\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:{y}_{{C}} =\:{y}_{\mathrm{1}} −{l}\mathrm{cos}\:\theta \\ $$$${And}\:{since}\:{C}\:{lies}\:{on}\:{ellipse} \\ $$$$\:\:\:\:\frac{\left({x}_{\mathrm{1}} −{l}\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}_{\mathrm{1}} −{l}\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\:\:\Rightarrow\:\:\frac{{l}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+\frac{{l}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }+\left(\frac{{x}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:−\mathrm{2}{l}\left(\frac{{x}_{\mathrm{1}} \mathrm{sin}\:\theta}{{a}^{\mathrm{2}} }+\frac{{y}_{\mathrm{1}} \mathrm{cos}\:\theta}{{b}^{\mathrm{2}} }\right)\:=\:\mathrm{1} \\ $$$${but}\:\:\left(\frac{{x}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\:=\:\mathrm{1},\:{So} \\ $$$$\Rightarrow\:\:{l}=\:\frac{\mathrm{2}\left(\frac{{x}_{\mathrm{1}} \mathrm{sin}\:\theta}{{a}^{\mathrm{2}} }+\frac{{y}_{\mathrm{1}} \mathrm{cos}\:\theta}{{b}^{\mathrm{2}} }\right)}{\frac{\mathrm{sin}\:^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }}\: \\ $$$$\:\:\:\bigtriangleup\:=\:\frac{{l}}{\mathrm{2}}\sqrt{{x}_{{A}} ^{\mathrm{2}} +{y}_{{A}} ^{\mathrm{2}} } \\ $$$$\:\:=\:\frac{\left(\frac{{x}_{\mathrm{1}} \mathrm{sin}\:\theta}{{a}^{\mathrm{2}} }+\frac{{y}_{\mathrm{1}} \mathrm{cos}\:\theta}{{b}^{\mathrm{2}} }\right)\sqrt{\frac{{a}^{\mathrm{4}} }{{x}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{{b}^{\mathrm{4}} }{{y}_{\mathrm{1}} ^{\mathrm{2}} }}}{\frac{\mathrm{sin}\:^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }}\: \\ $$$$\Rightarrow\:\bigtriangleup=\frac{\left(\frac{{x}_{\mathrm{1}} \mathrm{sin}\:\theta}{{y}_{\mathrm{1}} {a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:\theta}{{b}^{\mathrm{2}} }\right)\sqrt{\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} {a}^{\mathrm{4}} }{{x}_{\mathrm{1}} ^{\mathrm{2}} }+{b}^{\mathrm{4}} }}{\frac{\mathrm{sin}\:^{\mathrm{2}} \:\theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }} \\ $$$${using}\:\left({i}\right)\::\:\:\:\:\:\:\:\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }\:=\:\frac{{a}^{\mathrm{2}} \mathrm{sin}\:\theta}{{b}^{\mathrm{2}} \mathrm{cos}\:\theta}\: \\ $$$$\bigtriangleup\:=\:\frac{\left(\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} \mathrm{cos}\:\theta}+\frac{\mathrm{cos}\:\theta}{{b}^{\mathrm{2}} }\right)\sqrt{\frac{{b}^{\mathrm{4}} \mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:^{\mathrm{2}} \theta}+{b}^{\mathrm{4}} }}{\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }} \\ $$$$\:\:\bigtriangleup\:\:=\:\frac{\frac{\mathrm{1}}{{b}^{\mathrm{2}} \mathrm{cos}\:\theta}×\frac{{b}^{\mathrm{2}} }{\mathrm{sin}\:\theta}}{\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }}\: \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\bigtriangleup}\:=\:\mathrm{sin}\:\theta\mathrm{cos}\:\theta\left(\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{{a}^{\mathrm{2}} }+\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{{b}^{\mathrm{2}} }\right) \\ $$$${let}\:\:\:\:\:\mathrm{tan}\:\theta\:=\:{m} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{1}}{\bigtriangleup}\:=\:\frac{{m}\left(\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\frac{{d}\left(\mathrm{1}/\bigtriangleup\right)}{{d}\theta}\:=\:\frac{{d}\left(\mathrm{1}/\bigtriangleup\right)}{{dm}}×\frac{{dm}}{{d}\theta} \\ $$$$\frac{{d}\left(\mathrm{1}/\bigtriangleup\right)}{{d}\theta}\:=\:\mathrm{0}\:\:\left({for}\:{minimum}\:\bigtriangleup\right) \\ $$$$\Rightarrow\:\:\:\frac{{d}\left(\mathrm{1}/\bigtriangleup\right)}{{dm}}\:=\:\left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\frac{\mathrm{3}{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:−\:\mathrm{4}{m}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left(\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\left(\frac{\mathrm{3}{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)=\mathrm{4}{m}^{\mathrm{2}} \left(\frac{{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{3}{m}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{3}{m}^{\mathrm{4}} }{{a}^{\mathrm{2}} }+\frac{{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{m}^{\mathrm{4}} }{{a}^{\mathrm{2}} }+\frac{\mathrm{4}{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:{m}^{\mathrm{4}} +\mathrm{3}{m}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$${m}^{\mathrm{2}} \:=\:\sqrt{\frac{\mathrm{9}}{\mathrm{4}}\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}\:−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\mathrm{1}\right)\: \\ $$$$\bigtriangleup_{{min}} \:=\:\frac{{a}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)^{\mathrm{2}} }{{m}\left({m}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)}\:\:. \\ $$$$\left({As}\:{a}\:{special}\:{case}\right) \\ $$$${If}\:\:{a}=\:{b}\:={R}\:,\:\:{then}\:\:\:{m}=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\bigtriangleup_{{min}} =\:\mathrm{2}{R}^{\mathrm{2}} \:. \\ $$
Commented by mr W last updated on 11/Nov/18
great solution sir!
$${great}\:{solution}\:{sir}! \\ $$

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