判断平面内一个点 是否在两个点所确定的长轴椭圆内

    # 点到两个端点所确定的直线的距离　classic formula is:
    # # d = [(x2-x1)*(y1-y)-(x1-x)*(y2-y1)] / sqrt((x2-x1)**2 + (y2-y1)**2)
    #
    xD = (x2 - x1)
    yD = (y2 - y1)
    detaX = x1 - X
    detaY = y1 - Y
    norm2 = sqrt(xD ** 2 + yD ** 2)  # 注意norm2是一个数而不是numpy数组,因为xD, yD都是一个数。单个数字运算math比numpy快
    b = thre
    dist = xD * detaY - detaX * yD  # 常数与numpy数组(X,Y是坐标数组）的运算，broadcast
    dist /= norm2
    dist = np.abs(dist)
    ratiox = np.abs(detaX / (xD + 1e-8))
    ratioy = np.abs(detaY / (yD + 1e-8))
    ratio = np.where(ratiox < ratioy, ratiox, ratioy)
    ratio = np.where(ratio > 1, 1, ratio)  # 不用　np.ones_like(ratio)也可以正常运行，并且会快一点点
    ratio = np.where(ratio > 0.5, 1 - ratio, ratio)
    oncurve_dist = b * np.sqrt(1 - np.square(ratio * 2))  # oncurve_dist计算的是椭圆边界上的点到长轴的垂直距离

    sigma = thre  # todo: sigma of PAF 对于PAF的分布，设其标准差为多少最合适呢

    guass_dist[dist >= b] = 0

    return guass_dist