Question Number 48324 by mr W last updated on 22/Nov/18
Commented by ajfour last updated on 22/Nov/18
Commented by ajfour last updated on 22/Nov/18
![let OB = 2a eq. of OQ: y= mx eq. of BP: y = −s(x−2a) y_P = (7/a) , y_Q = (6/a) M(((2as)/(m+s)), ((2asm)/(m+s))) ⇒ (((2asm)/(m+s)))a = 4 ⇒ (a^2 /(((1/s)+(1/m)))) = 2 ⇒ (1/s)+(1/m) = (a^2 /2) .....(i) x_P = 2a−(7/(as)) ; x_Q = (6/(am)) eq.of OA: y = (((7/a)x)/(2a−(7/(as)))) eq. of AB: y= −(((6/a)(x−2a))/(2a−(6/(am)))) y_A = −(((6/a)[y_A (2a−(7/(as)))((a/7))−2a])/(2a−(6/(am)))) ⇒ y_A [(2a−(6/(am)))+(6/7)(2a−(7/(as)))] = 12 ⇒ y_A = ((12)/((((26a)/7)−(6/(am))−(6/(as))))) using (i) y_A = ((12)/(((26a)/7)−(6/a)((a^2 /2)))) = ((84)/(5a)) let Area(△AOB) = △ and required area = A, then △ = A+9 ay_A = A+9 ⇒ A = ((84)/5)−9 = ((39)/5) .](Q48332.png)
$${let}\:{OB}\:=\:\mathrm{2}{a}\: \\ $$$${eq}.\:{of}\:{OQ}:\:\:\:\:\:{y}=\:{mx} \\ $$$${eq}.\:{of}\:\:{BP}:\:\:\:\:{y}\:=\:−{s}\left({x}−\mathrm{2}{a}\right) \\ $$$${y}_{{P}} \:=\:\frac{\mathrm{7}}{{a}}\:,\:\:{y}_{{Q}} \:=\:\frac{\mathrm{6}}{{a}} \\ $$$${M}\left(\frac{\mathrm{2}{as}}{{m}+{s}},\:\frac{\mathrm{2}{asm}}{{m}+{s}}\right) \\ $$$$\Rightarrow\:\:\left(\frac{\mathrm{2}{asm}}{{m}+{s}}\right){a}\:=\:\mathrm{4}\:\:\:\:\:\:\:\: \\ $$$$\Rightarrow\:\:\frac{{a}^{\mathrm{2}} }{\left(\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{m}}\right)}\:=\:\mathrm{2}\: \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{m}}\:=\:\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:\:.....\left({i}\right) \\ $$$${x}_{{P}} \:=\:\mathrm{2}{a}−\frac{\mathrm{7}}{{as}}\:\:\:;\:\:{x}_{{Q}} \:=\:\frac{\mathrm{6}}{{am}} \\ $$$${eq}.{of}\:{OA}:\:\:\:\:{y}\:=\:\frac{\left(\mathrm{7}/{a}\right){x}}{\mathrm{2}{a}−\frac{\mathrm{7}}{{as}}} \\ $$$${eq}.\:{of}\:{AB}:\:\:\:{y}=\:−\frac{\left(\mathrm{6}/{a}\right)\left({x}−\mathrm{2}{a}\right)}{\mathrm{2}{a}−\frac{\mathrm{6}}{{am}}} \\ $$$$\:\:{y}_{{A}} \:=\:−\frac{\left(\mathrm{6}/{a}\right)\left[{y}_{{A}} \left(\mathrm{2}{a}−\frac{\mathrm{7}}{{as}}\right)\left(\frac{{a}}{\mathrm{7}}\right)−\mathrm{2}{a}\right]}{\mathrm{2}{a}−\frac{\mathrm{6}}{{am}}} \\ $$$$\Rightarrow\:\:{y}_{{A}} \left[\left(\mathrm{2}{a}−\frac{\mathrm{6}}{{am}}\right)+\frac{\mathrm{6}}{\mathrm{7}}\left(\mathrm{2}{a}−\frac{\mathrm{7}}{{as}}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{12} \\ $$$$\Rightarrow\:\:{y}_{{A}} \:=\:\frac{\mathrm{12}}{\left(\frac{\mathrm{26}{a}}{\mathrm{7}}−\frac{\mathrm{6}}{{am}}−\frac{\mathrm{6}}{{as}}\right)} \\ $$$${using}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:{y}_{{A}} \:=\:\frac{\mathrm{12}}{\frac{\mathrm{26}{a}}{\mathrm{7}}−\frac{\mathrm{6}}{{a}}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\right)}\:=\:\frac{\mathrm{84}}{\mathrm{5}{a}} \\ $$$$\:\:\:\:{let}\:\:{Area}\left(\bigtriangleup{AOB}\right)\:=\:\bigtriangleup \\ $$$${and}\:{required}\:{area}\:=\:{A},\:{then} \\ $$$$\:\:\:\bigtriangleup\:=\:{A}+\mathrm{9} \\ $$$$\:\:\:\:{ay}_{{A}} =\:{A}+\mathrm{9} \\ $$$$\Rightarrow\:\:\:\:{A}\:=\:\frac{\mathrm{84}}{\mathrm{5}}−\mathrm{9}\:=\:\frac{\mathrm{39}}{\mathrm{5}}\:. \\ $$
Commented by mr W last updated on 22/Nov/18
$${thanks}\:{for}\:{this}\:{analytical}\:{solution}! \\ $$
Commented by mr W last updated on 22/Nov/18
$${alternative}\:{way}: \\ $$$${let}\:{area}\:{of}\:{APMQA}={x} \\ $$$$\frac{{AQ}}{{QB}}=\frac{\mathrm{3}+{x}}{\mathrm{4}+\mathrm{2}}=\frac{\mathrm{3}+{x}}{\mathrm{6}} \\ $$$$\frac{{AP}}{{PO}}=\frac{\mathrm{2}+{x}}{\mathrm{4}+\mathrm{3}}=\frac{\mathrm{2}+{x}}{\mathrm{7}} \\ $$$$\frac{\Delta_{{AMQ}} }{\mathrm{2}}=\frac{{AQ}}{{QB}}=\frac{\mathrm{3}+{x}}{\mathrm{6}} \\ $$$$\Rightarrow\Delta_{{AMQ}} =\frac{\mathrm{3}+{x}}{\mathrm{3}} \\ $$$$\frac{\Delta_{{AMP}} }{\mathrm{3}}=\frac{{AP}}{{PO}}=\frac{\mathrm{2}+{x}}{\mathrm{7}} \\ $$$$\Rightarrow\Delta_{{AMP}} =\frac{\mathrm{3}\left(\mathrm{2}+{x}\right)}{\mathrm{7}} \\ $$$$\Delta_{{AMP}} +\Delta_{{AMQ}} ={x} \\ $$$$\Rightarrow\frac{\mathrm{3}\left(\mathrm{2}+{x}\right)}{\mathrm{7}}+\frac{\mathrm{3}+{x}}{\mathrm{3}}={x} \\ $$$$\Rightarrow\mathrm{9}\left(\mathrm{2}+{x}\right)+\mathrm{7}\left(\mathrm{3}+{x}\right)=\mathrm{21}{x} \\ $$$$\Rightarrow\mathrm{5}{x}=\mathrm{39} \\ $$$$\Rightarrow{x}=\mathrm{39}/\mathrm{5} \\ $$
Commented by ajfour last updated on 22/Nov/18
$${I}\:{dont}\:{think},\:{i}\:{can}\:{think}\:{like}\:{this}! \\ $$
Commented by mr W last updated on 22/Nov/18
$${you}'{re}\:{just}\:{too}\:{modest}\:{sir}! \\ $$