4、+Cy0+E)y+解析因为f(4,-8)=63,H=7(1-推论2过定点P(x0,y0)向双曲线22(Dx0+Ey0+F).y),由公式(1)得[7(1-y)]=63(x+xy+x2y2证明设Q(x,y)是经过点P(x0,y0)-=1引切线,当点P(x0,y)不在双y2-2y-1),a2b20的直线上的任一点,直线PQ与曲线Γ:即9x2+9xy+2y2-4y-16=0,分解因式曲线上时,切线方程为-x0x-y0y-1-2=f(x,y)=0交于点R(m,n),点R分 P Q所成a2b2(3x+y-4)(3x+2y+4)=0,x+λxx2y2x2y2所求切线方程为3x+y-4