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Question Number 2135 by Rasheed Soomro last updated on 04/Nov/15
Solve the following system of inequalities b^2 x^2 +a^2 y^2 ≤a^2 b^2 ∧ a^2 x^2 +b^2 y^2 ≤a^2 b^(2 ) ; a,b≠0
$${Solve}\:{the}\:{following}\:{system}\:{of}\:{inequalities} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} \:\leqslant{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\:\wedge\:\:\:{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} {y}^{\mathrm{2}} \:\leqslant{a}^{\mathrm{2}} {b}^{\mathrm{2}\:} \:\:;\:\:\:{a},{b}\neq\mathrm{0} \\ $$
Commented by 123456 last updated on 04/Nov/15
{ (((bx)^2 +(ay)^2 =(ab)^2 )),(((ax)^2 +(by)^2 =(ab)^2 )) :}
$$\begin{cases}{\left({bx}\right)^{\mathrm{2}} +\left({ay}\right)^{\mathrm{2}} =\left({ab}\right)^{\mathrm{2}} }\\{\left({ax}\right)^{\mathrm{2}} +\left({by}\right)^{\mathrm{2}} =\left({ab}\right)^{\mathrm{2}} }\end{cases} \\ $$
Answered by prakash jain last updated on 04/Nov/15
x^2 =u,y^2 =v,a^2 =c,b^2 =d Also assume c≤d du+cv=cd cu+dv=cd u=((cd)/(c+d)),v=((cd)/(c+d)) u=0, v≤c v=0, u≤c Range of u,v or x^2 ,y^2 will be inside the quadilateral (0,0),(0,c),(((cd)/(c+d)),((cd)/(c+d))),(c,0) This mapped to x,y will be within the octagon with the following points assuming a^2 ≤b^2 . (−a,0),(((−ab)/( (√(a^2 +b^2 )))),((−ab)/( (√(a^2 +b^2 ))))),(0,−a),(((ab)/( (√(a^2 +b^2 )))),((−ab)/( (√(a^2 +b^2 ))))), (a,0),(((ab)/( (√(a^2 +b^2 )))),((ab)/( (√(a^2 +b^2 ))))),(0,a),(((−ab)/( (√(a^2 +b^2 )))),((ab)/( (√(a^2 +b^2 ))))),
$${x}^{\mathrm{2}} ={u},{y}^{\mathrm{2}} ={v},{a}^{\mathrm{2}} ={c},{b}^{\mathrm{2}} ={d} \\ $$$$\mathrm{Also}\:\mathrm{assume}\:{c}\leqslant{d} \\ $$$${du}+{cv}={cd} \\ $$$${cu}+{dv}={cd} \\ $$$${u}=\frac{{cd}}{{c}+{d}},{v}=\frac{{cd}}{{c}+{d}} \\ $$$${u}=\mathrm{0},\:{v}\leqslant{c} \\ $$$${v}=\mathrm{0},\:{u}\leqslant{c} \\ $$$$\mathrm{Range}\:\mathrm{of}\:{u},{v}\:\mathrm{or}\:{x}^{\mathrm{2}} ,{y}^{\mathrm{2}} \mathrm{will}\:\mathrm{be}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{quadilateral} \\ $$$$\left(\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},{c}\right),\left(\frac{{cd}}{{c}+{d}},\frac{{cd}}{{c}+{d}}\right),\left({c},\mathrm{0}\right) \\ $$$$\mathrm{This}\:\mathrm{mapped}\:\mathrm{to}\:{x},{y}\:\mathrm{will}\:\mathrm{be}\:\mathrm{within}\:\mathrm{the}\:\mathrm{octagon} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{following}\:\mathrm{points}\:\mathrm{assuming}\:{a}^{\mathrm{2}} \leqslant{b}^{\mathrm{2}} . \\ $$$$\left(−{a},\mathrm{0}\right),\left(\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right),\left(\mathrm{0},−{a}\right),\left(\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right), \\ $$$$\left({a},\mathrm{0}\right),\left(\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right),\left(\mathrm{0},{a}\right),\left(\frac{−{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\right), \\ $$

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