在直角坐标系xOy中,动点P到两点(0,-√3),(0,√3)的距离之和等于4,设点P的轨迹为曲线C,直线y＝kx＋1与

1为x^2/a^2+y^2/b^2=1,则2a=4,c=√3,解x^2/4+y^2=1.
2.(1+4k^2)x^2+8KX=0
x1+x2=(1+4k^2)/8k,x1*x2=0.
d^2=(x1-x2)^2=(x1+x2)^2-4x1x2=(1+4k^2/8)^2即d=1+4K^2/8k,
h=1/1+k2,则S=1/2*d*h,0