When y=0, x=- 1 or 3, so the coordinate of point A is (-1, 0).

At this time, the AC distance is 10, and the AC midpoint coordinate is (-1/2,3/2).

The vertical line of AC is x+3y-4=0,

Finding the intersection of the median vertical line and parabola will make ANC an isosceles triangle.

Then let the coordinates of point n be (x, y)

Then x 2+(y-3) 2 = 10 (cn distance =AC).

Because the distance from point C to the vertex of parabola = root number 2 is smaller than the radius of the circle, the circle X 2+(y-3) 2 =10 has only two intersections with parabola, one of which is (-1, 0).

Otherwise, (x+1) 2+y 2 =10 (an distance =AC). At this time, the circle (x+ 1) 2+y 2 = 10 has two intersections with the parabola, one of which is (0,3) inconsistent.

There are four points that can make ANC an isosceles triangle.