Question

Помогите, пожалуйста, решить. С подробным решением

Answer 1

$\frac{x}{y}=m\Rightarrow\frac{y}{x}=\frac{1}{m} \\\\m-\frac{1}{m}=\frac{8}{3} \\\\m-\frac{1}{m}-\frac{8}{3}=0\\\\\frac{3m^{2}-8m-3 }{3m}=0\\\\\left \{ {{3m^{2}-8m-3=0 } \atop {m\neq 0 }} \right.\\\\D=(-8)^{2}-4*3*(-3)=64+36=100=10^{2}\\\\m_{1}=\frac{8+10}{6} = 3\\\\m_{2}=\frac{8-10}{6}=-\frac{1}{3}$

1)

$\left \{ {{\frac{x}{y}=3 } \atop {x^{2}+y^{2}=360}} \right.\\\\\left \{ {{x=3y} \atop {9y^{2}+y^{2}=360}} \right.\\\\\left \{ {{x=3y} \atop {10y^{2}=360}} \right.\\\\\left \{ {{x=3y} \atop {y^{2}=36 }} \right.\\\\\left \{ {{x_{1}=18} \atop {y_{1}=6}} \right.\\\\\left \{ {{x_{2}=-18} \atop {y_{2}=-6 }} \right.$

2)

$\left \{ {{\frac{x}{y}=-\frac{1}{3}} \atop {x^{2}+y^{2}=360}} \right.\\\\\left \{ {{y=-3x} \atop {x^{2}+(-3x)^{2}=360}} \right.\\\\\left \{ {{y=-3x} \atop {x^{2}+9x^{2}=360}} \right. \\\\\left \{ {{y=-3x} \atop {10x^{2}=360}} \right.\\\\\left \{ {{y=-3x} \atop {x^{2}=36}} \right.\\\\\left \{ {{x_{3}=6 } \atop {y_{3}=-18}} \right.\\\\\left \{ {{x_{4}=-6 } \atop {y_{4}=18 }} \right.$

Ответ :(18 ; 6), (- 18 ; - 6), (6 ; - 18),(- 6; 18)