Question

زوج های مرتب $$(x,y)$$ را طوری بیابید که: $$ x^{2} = \frac{10}{x+9y} \wedge y^{2} = \frac{2}
{x+y} $$

Answer 1

$$x=rsin \alpha \wedge y=rcos \alpha \wedge \alpha not equal \frac{k \pi }{2} \Longrightarrow \frac{ x^{2} }{ y^{2} } = \frac{5x+5y}{x+9y} \Longrightarrow \frac{ sin^{2} \alpha }{ cos^{2 } \alpha }= \frac{5sin \alpha +5cos \alpha }{sin \alpha +9cos \alpha } \Longrightarrow sin^{3} \alpha +9 sin^{2} \alpha cos \alpha =5sin \alpha cos^{2} \alpha +5 cos^{3} \alpha \Longrightarrow ÷cos \alpha is not zero \Longrightarrow tan^{3} \alpha +9 tan^{2} \alpha -5tan \alpha -5=0 \Longrightarrow (tan \alpha -1)( tan^{2} \alpha +10tan \beta +5)=0 \Longrightarrow tan \alpha =1 \Longrightarrow \frac{x}{y} =1 \Longrightarrow x^{3} =1 \Longrightarrow x=y=1 \Longrightarrow (x,y)=(1,1) \vee tan^{2} \alpha +10tan \alpha +5=0 \Longrightarrow tan \alpha =-5 \pm 2 \sqrt{5} \Longrightarrow i) \frac{x}{y} =-5+2 \sqrt{5} \Longrightarrow \frac{x+y}{y} =2 \sqrt{5} -4 \Longrightarrow y^{3} = \frac{1}{ \sqrt{5} -2} \Longrightarrow y^{3} = \sqrt{5} +2 \Longrightarrow y= \sqrt[3]{ \sqrt{5} +2} \vee x=(2 \sqrt{5} -5). \sqrt[3]{ \sqrt{5} +2} \Longrightarrow (x,y)=((2 \sqrt{5} -5). \sqrt[3]{ \sqrt{5}+2 } , \sqrt[3]{ \sqrt{5}+2 } ) \vee ii) \frac{x}{y} =-5-2 \sqrt{5} \Longrightarrow \frac{x+y}{y} =-4-2 \sqrt{5} \Longrightarrow y^{3} = \frac{-1}{ \sqrt{5} +2} \Longrightarrow y^{3} =2- \sqrt{5} \Longrightarrow y= \sqrt[3]{2- \sqrt{5} } \vee x=(-5-2 \sqrt{5} ). \sqrt[3]{2- \sqrt{5} } \Longrightarrow (x,y)=((-5-2 \sqrt{5} ). \sqrt[3]{2- \sqrt{5} } , \sqrt[3]{2- \sqrt{5} }) $$