直線y=ax+1和雙曲線3x^2-y^2=1交于Ａ，Ｂ兩點，a為何值時，以Ａ，Ｂ為直徑的圓經過原點？

熱心網友

直線y=ax+1和雙曲線3x^2-y^2=1交于Ａ，Ｂ兩點，a為何值時，以Ａ，Ｂ為直徑的圓經過原點？　 解:設A(x1,y1) B(x2,y2) 圓心坐標(x,y)因為: Ａ,Ｂ為直徑的圓經過原點(OA垂直OB)所以: x1*x2=-y1*y2y=ax+1。。。。。。。。。(1)3x^2-y^2=1。。。。。(2)(1)代入(2):(3-a^2)x^2-2ax-2=0x1+x2=2a/(3-a^2)=====x=a/(3-a^2)x1*x2=-2/(3-a^2)同理:(3-a^2)y-6y+(3-a^2)=0y1+y2=6/(3-a^2)======y=3/(3-a^2)y1*y2=1所以:-2/(3-a^2)=-1a^2=1 ===============a1=1 a2=-1所以:圓心1的坐標(1/2,3/2) 圓心2的坐標(-1/2),3/2)半徑R^2=(1/2)^2+(3/2)^2=5/2所以:園的方程為:(2x-1)^2+(2y-3)^2=10 或 (2x+1)^2+(2y-3)^2=10。

熱心網友

OA、OB垂直時，以A(x1,y1)、B(x2,y2)為直徑的圓經過原點即：(y1/x1)*(y2/x2) = -1 === x1*x2 +y1*y2 = 0將y=ax+1代入3x^2-y^2=1：(a^2-3)x^2 +2ax+2 = 0x1*x2 +y1*y2 = x1*x2 +(a*x1+1)(a*x2+1) = (1 +a^2)x1*x2 +2a*(x1+x2)+1 = 0== (1 +a^2)*[2/(a^2 -3)] + a*[(-2a)/(a^2 -3)]x1 + 1 = 0== a = 1 or -1