Question

当m取何值时,方程组x^2+2y^2-6=0 y=mx+3,这时解为?

guichen

Answer 1

设x=Sqrt(6)*cos(t)，y=Sqrt(3)*sin(t)
y=mx+3 =>
Sqrt(3)*sin(t)=m*(Sqrt(6)*cos(t))+3
=>3*sin^2(t)=6*m^2*cos^2(t)+6*Sqrt(6)*cos(t)+9
=>(1+2*m^2)*sin^2(t)+2*Sqrt(6)*m*sin(t)+2=0
=>sin(t)=-sqrt(6)*m加减sqrt(2*(m^2-1))/(1+2*m^2)
由于-1<=sin(t)<=1
=>m<=-1或m>=1
方程组x^2+2y^2-6=0 y=mx+3 联解出
x=2*(-3*m+Sqrt(3*(m^2-1)))/(2*m^2+1)
y=(3+2*m*Sqrt(3*(m^2-1)))/(2*m^2+1)
或
x=2*(-3*m-Sqrt(3*(m^2-1)))/(2*m^2+1)
y=(3-2*m*Sqrt(3*(m^2-1)))/(2*m^2+1)

Answer 2

联立x^2+2y^2-6=0
y=mx+3
得(1+m^2)x^2+6mx+3=0
△=36m^2-12-12m^2
=24m^2-12
△=0有一个解
m=±√2/2,x=±√3
△>0有两个解

Answer 3

-12m+根号(12*m*12*m-48*(2*m*m+1))或
-12m-根号(12*m*12*m-48*(2*m*m+1))