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I have been trying to find the angle made by inner tangents to two general ellipse by following the method described by Futurologist to find four homogeneous equations of tangents, this method is presented here and uses the concept of dual conics and degenerate conics. Using that method, I am able to get four tangent equations in homogeneous coordinates. However, I am looking to find the angle between the inner common tangents. For that purpose, currently, I am solving for the intersection of the four tangent lines with the ellipse and then tracing those points back to get the angle made by inner common tangents. This process is a bit computationally expensive and I am looking for a less expensive algorithm.

My question here is that if it is possible to obtain that angle directly instead of going through the hassle of finding the intersection points? I am not interested in getting the equation of tangents. I just need the angle between them.

Graphically in the diagram below, that angle can be represented as $\theta$:

Note: Using the slopes of the lines obtained from above method do not return correct answer because we do not know if that angle is greater one or smaller one (two angle made by exterior or interior to those line), it depends on the relative position of the two conics. I want a generalize solution which will work for every configuration of two given conics anywhere in the 2D space.

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  • $\begingroup$ Can you specify what is the way of defining the conics? Are they given in the form of two quadratic equations of two variables, or as position of center, angle between major axis and a coordinate axis and magnitudes of major and minor axes of each conic? Or maybe some other way? Are the conics always ellipses or circles? If not, then what is considered internal common tangents when you have hyperbolas and parabolas involved? $\endgroup$ Commented Feb 14, 2021 at 19:12
  • $\begingroup$ @Futurologist, the conics are given in the same form as in the previous thread I referred. I wanted to get the angle between internal common tangents between any two general conics but if you think it is not as clear for hyperbolas and parabolas, for now, let's consider conics are always ellipse as in the previous thread. $\endgroup$ Commented Feb 14, 2021 at 23:28
  • $\begingroup$ @Futurologist, I have updated my question with a link to definition of general ellipse given in the quadratic form. $\endgroup$ Commented Feb 15, 2021 at 0:58

2 Answers 2

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Preliminaries (algorithm for solving a pair of quadratic equations of two variables):

It boils down to finding an algorithm for solving a system of two quadratic equations of two variables, by interpreting it as a projective geometry pencil of conics, then finding the three degenerate conics of the pencil together with a projective transformation that simplifies these three degenerate conics to the point of your being able to read off the solutions very easily in new simplifying coordinate system, and after that transforming them back to the original coordinate system.

I sketched an algorithm in python, I think it seems to work on your example... but I was in a hurry and did not check it properly, so there may be bugs...

import numpy as np
import math

# technical module, functions, details
def homogenize(x):
    return np.array([x[0], x[1], 1])

def cos_sin(angle_deg):
    return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)

def rotation(cs_sn):
    return np.array([[cs_sn[0], -cs_sn[1]], 
                     [cs_sn[1],  cs_sn[0]]])

# defining the isometry (the rotation plus translation) transformation
# between the coordinate system aligned with the conic and a general (world) 
# coordinate system
def isom_inverse(angle, translation):
    '''
    isometry from conic-aligned coordinate system (conic attached)
    to global coordinate system (world system) 
    '''
    cos_, sin_ = cos_sin(angle)
    return np.array([[cos_, -sin_, translation[0]], 
                     [sin_,  cos_, translation[1]], 
                     [   0,     0,             1]])
    
def isom(angle, translation):
    '''
    isometry from global coordinate system (world system) 
    to conic-aligned coordinate system (conic attached) 
    '''
    cos_, sin_ = cos_sin(-angle)
    tr = - rotation((cos_, sin_)).dot(translation)
    return np.array([[ cos_, -sin_, tr[0]], 
                     [ sin_,  cos_, tr[1]], 
                     [    0,     0,    1 ]])

# calculating the coinc defined by a pair of axes' lengts, 
# axes rotation angle and center of the conic  
def Conic(major, minor, angle, center):
    D = np.array([[minor**2,        0,                 0],
                  [       0, major**2,                 0], 
                  [       0,        0, -(minor*major)**2]])
    U = isom(angle, center)
    return (U.T).dot(D.dot(U))     

# calculating the coinc dual to the conic defined by a pair of axes' lengths, 
# axes rotation angle and center of the conic  
def dual_Conic(major, minor, angle, center):
    D_1 = np.array([[major**2,        0,  0], 
                    [       0, minor**2,  0], 
                    [       0,        0, -1]])
    U_1 =  isom_inverse(angle, center)
    return (U_1).dot(D_1.dot(U_1.T)) 

# transforming the matrix of a conic into a vector of six coefficients
# of a quadratic equation with two variables
def conic_to_equation(C):
    '''
    c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
    '''
    return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])    

# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of 
# the corresponding conic 
def equation_to_conic(eq):
    '''
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

# given a point (x,y) define the vector (x^2, xy, y^2, x, y, 1)    
def argument(x):
    return np.array([x[0]**2, x[0]*x[1], x[1]**2, x[0], x[1], 1])

# given x = (x[0],x[1]) calculate the value of the quadratic equation with
# six coefficients coeff
def quadratic_equation(x, coeff):
    '''
    coeff[0]*x**2 + coeff[1]*x*y + coeff[2]*y**2 + coeff[3]*x + coeff[4]*y + coeff[5] = 0
    '''
    return coeff.dot( argument(x) )    

# given a pair of conics, as a pair of symmetric matrices, 
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which 
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2 
# is a degenerate conic (the anti-symmetric product of a pair of linear forms) 
# and also find the matrix U 
# of the projective transformation that simplifies the geometry of 
# the pair of conics, the geometry of the pencil c1 - t*c2 in general, 
# as well as the geometry of the three degenerate conics in particular    
def transform(c1, c2):
    '''
    c1 and c2 are 3 by 3 symmetric matrices of the two conics
    '''
    c21 = np.linalg.inv(c2).dot(c1)
    k, U = np.linalg.eig(c21)
    return k, U

# the same as before, but for a pair of equations instead of matrices of conics
def eq_transform(eq1, eq2):
    '''
    eq1 and eq2 = np.array([eq[0], eq[1], eq[2], eq[3], eq[4], eq[5]])
    '''
    C1 = equation_to_conic(eq1)
    C2 = equation_to_conic(eq2)
    return transform(C1, C2)

# realizing the matrix U as a projective transformation
def proj(U, x):
    if len(x) == 2:
        x = homogenize(x)
    y = U.dot(x)
    y = y / y[2]
    return y[0:2]

# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_common_points(c1, c2):
    k, U = transform(c1, c2)
    L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
    L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
    sol = np.empty((4,3), dtype=float)
    for i in range(2):
        for j in range(2):
            sol[i+2*j,0:2] = np.array([math.sqrt(abs(L2[2,2] / L2[0,0]))*(-1)**i, math.sqrt(abs(L1[2,2] / L1[1,1]))*(-1)**j])
            sol[i+2*j,0:2] = proj(U, sol[i+2*j,0:2])
    sol[:,2] = np.ones(4)
    return sol

# find the solutions, i.e. the points x=(x[0],x[1]) saisfying the pair 
# of quadratic equations 
# represented by a pair of vectors eq1 and eq2 of 6 coefficients
def solve_eq(eq1, eq2):
    conic1 = equation_to_conic(eq1)
    conic2 = equation_to_conic(eq2)
    return find_common_points(conic1, conic2)


'''
Esample of finding the common tangents of a pair of conics:
conic 1: major axis = 2, minor axis = 1, angle = 45, center = (0,0)
conic 2: major axis = 3, minor axis = 1, angle = 120, center = (15,0)
'''

a = 2
b = 1
cntr = np.array([0,0])
w = 45

Q1 = Conic(a, b, w, cntr)
dQ1 = dual_Conic(a, b, w, cntr)

a = 3
b = 1
cntr = np.array([15,0])
w = 120

Q2 = Conic(a, b, w, cntr)
dQ2 = dual_Conic(a, b, w, cntr)

R = find_common_points(dQ1, dQ2)

print('')
print(R)
print('')
print('checking that the output forms common tangent lines: ')
print('')
print('conic 1: ')
print(np.diagonal(R.dot(dQ1.dot(R.T))) )
print('')
print('conic 2: ')
print(np.diagonal(R.dot(dQ2.dot(R.T))) )
#conic_to_equation(dQ1)

Some Explanations: Assume you want to find the intersection points of two conics C1 and C2. Let us assume for simplicity that they are real ellipses that intersect in four different points (to avoid complex numbers)

In the case of finding the common tangents to a pair of conics, simply convert the two conics two corresponding duals and then find the intersection points of the duals. These intersection points are the equation coefficients of the tangents of the original conics.

There are possibly several different geometric interpretations of this problem, but let us go with the pencil of conics. The two conics C1 and C2 are represented by 3 by 3 symmetric matrices with non-zero determinants, which I have denoted C1 and C2. The linear combination, called pencil of conics generated by C1 and C2, is the t-parametrized family of conics C1 - t*C2 , where t is just a number. What is crucial is that every conic C1 - tC2 passes through the intersection points of C1 and C2 and these are the only four points they all have in common. You can prove this by observing that if x.T * C1 * x = x.T * C1 * x = 0 then x.T * (C1 - t*C2) * x = x.T * C1 * x - t * x.T * C2 * x = 0 - t*0 = 0. Furthermore, if you take an intersection point of C1 and C1 - t*C2, then C2 = C1 - t*C2 + s*C2 you can apply the same argument when s = t.

In this family of conics, there are three degenerate conics, that are geometrically three pairs of lines. They occur exactly when t is such that det( C1 - t*C2 ) = 0. This is a polynomial equation of degree 3 with respect to t, so there are three, say different solutions k[0], k[1], k[2], due to the niceness of the C1 and C2. Projectively speaking, each degenerate conic C1 - k[j]*C2 is a pair of lines and they have a common intersection point u[:,j] = [ u[0,j] : u[1,j] : u[2,j] ]. Moreover, rank(C1 - k[j]*C2) = 2, so ker(C1 - k[j]*C2) = 1. This point u[:,j] is characterized as a solution to the equation (C1 - k[j]*C2) * u[:,j] = 0. Since C2 is invertible (non-degenerate), multiply both sides of the equation by inverse(C2) and obtain the equivalent equation ( (inverse(C2) * C1) - k[j]*Identity ) * u[:,j] = 0 which is an eigenvalue equation, with k[j] as eigenvalue and u[:,j] as eigenvector. The output of the function transform() is the 1 by 3 array k of eigenvalues and the 3 by 3 matrix U = [ u[:,0], u[:,1], u[:,2] ] of eigenvectors.

Conjugating C1 - k[j]*C2 by U, i.e. (U.T)*(C1 - k[j]*C2)*U, is geometrically equivalent to performing a projective transformation that sends u[:,0] and u[:,1] to infinity and u[:,2] to the origin. Thus, U changes the coordinate system from a general one to a special coordinate system, in which two of the degenerate conics are given by pairs of parallel lines and combined they intersect in a rectangle. The third conic is represeneted by the diagnols of the rectangle.

In this new picture, the intersection problem can be easily solved, just by reading off one solution from the entries of the matrices (U.T)*(C1 - k[j]*C2)*U (the intersection points are the vertices of the rectangle, so if you find one of them, the others are simply mirror symmetric of each other).

Some explanations on how the solutions in special coordinates are obtained.

Since, by construction (by the special choice of eigenvalues k[0] and k[1]) the original conics C1 - k[0]*C2 and C1 - k[1]*C2 are degenerate and thus a pair of lines each, after the projective transformation U, which simplifies the conics and represents them by the matrices L1 = (U.T)*(C1 - k[0]*C2)*U and L1 = (U.T)*(C1 - k[0]*C2)*U, are diagonal and thus geometrically each of these is a pair of lines, parallel one of the coordinate axes, e.g. L1 is a pair of lines parallel to x and L2 is a pair of lines parallel to y. Also for each degenerate conic, the pair of lines are symmetric with respect to the coordinate axes. Hence, the four intersection points form the vertices of a rectangle, mirror symmetric relative to the coordinate axes. Thus, the intersection points are of the projective form

sol[0, : ] = [ sol[0, 0],  sol[0, 1],  1]
sol[1, : ] = [-sol[0, 0],  sol[0, 1],  1]
sol[2, : ] = [-sol[0, 0], -sol[0, 1],  1]
sol[3, : ] = [ sol[0, 0], -sol[0, 1],  1]

so it is enough to find only one of them and then reflect it across the coordinate axes. In the simplifying coordinate system, the conic L1 has the an equation like this

(a1*y - b1) * (a1*y + b1) = 0

where the two lines are a1*y = b1 and a1*y = -b1 so the y coordinate of one of the solutions should be y = abs(b1/a1). If you expand the equation, you get the quadratic equation

a1^2 * y^2 - b1^2 = 0

and in matrix form

       0     0      0
L1 =   0  L1[1,1]   0
       0     0   L1[2,2]

where   L1[1,1] = a1^2   and   L1[2,2] = b1^2 

Hence, the y coordinate of one solution is

y = abs(b1/a1) = sqrt( abs( L1[2, 2] / L1[1, 1] ) )

If you do the same for the conic L2, but this time thinking about it being a pair of symmetric lines parallel to the y axis, then you get the x coordinate of a solution

x = abs(b2/a2) = sqrt( abs( L2[2, 2] / L2[1, 1] ) )

That is why, one solution is

sol[0, :] = [ sqrt(abs( L2[2, 2] / L2[1, 1] )),  sqrt(abs( L1[2, 2] / L1[1, 1] )),  1]

Then all solutions are obtained by reflecting in the coordinate axes:

sol[, :] = [(-1)**i*sqrt(abs( L2[2, 2] / L2[1, 1] )),  (-1)**j*sqrt(abs( L1[2, 2] / L1[1, 1] )),  1]

for i = 0,1 and j=0,1

Algorithm for calculating the angle between the internal tangents:

import numpy as np
import math

'''
####### FUNCTIONS IMPORTANT FOR THE CALCULATION #######
'''
    
def homogenize(x):
    return np.array([x[0], x[1], 1])

# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of 
# the corresponding conic 
def conic_from_equation(eq):
    '''
    eq is an np.array vector of the equadratic equation's six coefficients
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    conc is a symmetric matrix of the bilinier form of the quadratic equation
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

# obtaining the conic dual to a given conic, 
# both represented as symmetric 3x3 matrices
def dual_conic(conic):
    return np.linalg.inv(conic)

# given a pair of conics, as a pair of symmetric 3x3 matrices, 
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which 
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2 
# is a degenerate conic (the anti-symmetric product of a pair of linear forms) 
# and also find the matrix U 
# of the projective transformation that simplifies the geometry of 
# the pair of conics, the geometry of the pencil c1 - t*c2 in general, 
# as well as the geometry of the three degenerate conics in particular  
def trivialization_of(conic1, conic2):
    '''
    c1 and c2 are 3 by 3 symmetric matrices of two conics (possibly dual)
    '''
    c21 = np.linalg.inv(conic2).dot(conic1)
    k, U = np.linalg.eig(c21)
    return k, U     

# realze the 3x3 matrix U as a projective transformation of projective 2-plane
def projectivity(U, x):
    if len(x) == 2:
        x = homogenize(x)
    y = U.dot(x)
    y = y / y[2]
    return y[0:2]

# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_intersection_points(c1, c2):
    k, U = trivialization_of(c1, c2)
    L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
    L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
    sol = np.empty((4,3), dtype=float)
    for i in range(2):
        for j in range(2):
            sol[i+2*j,0:2] = np.array([math.sqrt(abs(L2[2,2] / L2[0,0]))*(-1)**i, math.sqrt(abs(L1[2,2] / L1[1,1]))*(-1)**j])
            sol[i+2*j,0:2] = projectivity(U, sol[i+2*j,0:2])
    sol[:,2] = np.ones(4)
    return sol    

# find the three coefficients of all four common tangents to a pair of conics
# make sure that the conics do have four common tangents, 
# which is the generic situation
def find_common_tangents(c1, c2):
    dc1 = dual_conic(c1)
    dc2 = dual_conic(c2)
    return find_intersection_points(dc1, dc2)

# if the two conics are ellipses, they do not intersect the line at infinity.
# by polarity, the poles of the line at infinity are 
# the (affine and in fact metric) centers of the ellipses and are located
# inside the ellipses each. We use them to detect the inner tangents and 
# to orient the vectors normal to these tangents in a way 
# so that the angle between these two normal vectors is the vector on the picture    
def dual_to_infinity_line(conic):
    p = dual_conic(conic).dot(np.array([0,0,1]))
    return homogenize( p[0:2] / p[2] )

# find the common inner tangents between a pair of conics, 
# represented as 3x3 symmetric matrices, with properly oriented normal vectors
def find_common_inner_tangents(c1, c2):
    p1 = dual_to_infinity_line(c1)
    p2 = dual_to_infinity_line(c2)
    tn = find_common_tangents(c1, c2)
    # boolean vector, true if inner tangent false if not
    is_inner_tangent = ( tn.dot(p1)*tn.dot(p2) < 0 )
    tn = tn[ is_inner_tangent, : ]
    # reorient one of the normal vectors so that they give us the proper angle
    if tn[0,:].dot(p1)*tn[1,:].dot(p1) > 0:
        tn[1,:] = - tn[1,:]
    return tn

# calculate the angle between a pair of line, given by their three coefficients
def angle_bn_lines(line1, line2):
    theta = line1[0:2].dot(line2[0:2]) / math.sqrt( line1[0:2].dot(line1[0:2])*line2[0:2].dot(line2[0:2]))
    return math.acos(theta)*180/math.pi

# calculate the proper angle between the inner tangents of two quadratic equations
# that represent a pair of conics, hopefully non-intersecting
def angle_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
    c1 = conic_from_equation(eq_of_conic1)
    c2 = conic_from_equation(eq_of_conic2)
    tn = find_common_inner_tangents(c1, c2)
    return angle_bn_lines(tn[0,:], tn[1,:])
    
'''
####### END OF FUCTIONS IMPORTANT FOR THE CALCULATION #######
'''


'''
####### auxiliary functions to handle the test scenario: 
'''

def cos_sin(angle_deg):
    return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)

def rotation(cs_sn):
    return np.array([[cs_sn[0], -cs_sn[1]], 
                     [cs_sn[1],  cs_sn[0]]])
    
def isom(angle, translation):
    '''
    isometry from global coordinate system (world system) 
    to conic-aligned coordinate system (conic attached) 
    '''
    cos_, sin_ = cos_sin(-angle)
    tr = - rotation((cos_, sin_)).dot(translation)
    return np.array([[ cos_, -sin_, tr[0]], 
                     [ sin_,  cos_, tr[1]], 
                     [    0,     0,    1 ]])

# calculating the conic defined by a pair of axes' lengths, 
# axes rotation angle and center of the conic  
def Conic(major, minor, angle, center):
    D = np.array([[minor**2,        0,                 0],
                  [       0, major**2,                 0], 
                  [       0,        0, -(minor*major)**2]])
    U = isom(angle, center)
    return (U.T).dot(D.dot(U))     

# transfromaing the matrix of a conic into a vector of six coeficients
# of a quadratic equation with two variables
def equation_from_conic(C):
    '''
    c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
    '''
    return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])    
'''
####### end of auxiliary functions
'''

'''
Example of finding the angle between the common tangents of a pair of conics:
conic 1: major axis = 2, minor axis = 1, angle = 45, center = (0,0)
conic 2: major axis = 3, minor axis = 1, angle = 120, center = (15,0)
'''

a = 2
b = 1
cntr = np.array([0,0])
w = 45

Eq1 = equation_from_conic(Conic(a, b, w, cntr))
Co1 = conic_from_equation(Eq1)

a = 3
b = 1
cntr = np.array([15,0])
w = 120

Eq2 = equation_from_conic(Conic(a, b, w, cntr))
Co2 = conic_from_equation(Eq2)

four_tngs = find_common_tangents(Co1, Co2)

inner_tngs = find_common_inner_tangents(Co1, Co2) 

print( angle_bn_inner_tangents_of(Eq1, Eq2) )

Calculating the angle bisector line of the inner tangents.

Here is a routine that, given a pair of lines line1 and line2, represented by vectors with three coefficients, calculates the three coefficients of the angle bisector line that splits the angle in half. It also returns the slope of the tangent line, as the ratio of the y-increase divided by the x-increase.

def angle_bisector(line1, line2):
    n1 = line1[0:2]
    n2 = line2[0:2]
    n1 = n1 / math.sqrt(n1.dot(n1))
    n2 = n2 / math.sqrt(n2.dot(n2))
    n_bsec = n1 + n2
    p12 = np.cross(line1, line2)
    p12 = p12[0:2] / p12[2]
    c_bsec = n_bsec.dot(p12) 
    return np.array([n_bsec[0], n_bsec[1], c_bsec]), np.array([-n_bsec[1], n_bsec[0]])

def angle_and_bsec_vector_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
    c1 = conic_from_equation(eq_of_conic1)
    c2 = conic_from_equation(eq_of_conic2)
    tn = find_common_inner_tangents(c1, c2)
    if abs(np.all(tn.imag)) < 1e-10:
        tn = tn.real
    p1 = dual_to_infinity_line(c1)[0:2]
    p2 = dual_to_infinity_line(c2)[0:2]
    bsec, v_bsec = angle_bisector(tn[0], tn[1])
    angle = angle_bn_C_lines(tn[0,:], tn[1,:])
    if abs(angle.imag) < 1e-10:
        angle = angle.real
    if v_bsec.dot(p2 - p1) < 0:
        v_bsec = - v_bsec
    return angle, v_bsec / np.linalg.norm(v_bsec)
$\endgroup$
11
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Aug 15, 2021 at 14:58
  • $\begingroup$ Hello @Futurologist, is it also possible to get the equation and slope of angular bisector line which bisects the two inner common tangents? When I try to calculate the slope of that line using equations of inner common tangents we obtained then sometimes it differs by 180 degrees. Do you think I should post a new question for that? I just need it for the real case not the imaginary one. $\endgroup$ Commented Oct 19, 2021 at 22:20
  • 1
    $\begingroup$ Hello @KashishDhal I added a function at the end of the post. I hope it is what you are looking for. And maybe you should include me as a collaborator to your project. $\endgroup$ Commented Oct 20, 2021 at 3:31
  • $\begingroup$ Hello @Futurologist, sure we can add you as a collaborator in our next research paper. I will ask for your details prior to sending next paper. Also, I think you should have minus sign in the n_bsec[0]/n_bsec[1] term to get the slope. However, irrespective of that, I do not get correct slope all the time. Sometime the slope is shifted by angle of 180 degrees. Also, I found another thread here: math.stackexchange.com/a/3216295/861640 , can we use the center of the ellipse to detect the correct sector in which the angular bisector would be located? I tried but that didn't work. $\endgroup$ Commented Oct 20, 2021 at 4:58
  • $\begingroup$ @KashishDhal I do not understand the issue you are having. A line rotated 180 degrees is the same line. Consequently the slope should be the same after 180 degree rotation. The previous code already makes sure that the normal vectors of the two inner tangent lines are oriented so that the angle between them is equal to the angle between the inner tangents measured in the sectors containing the ellipses. Hence, the angle bisector's normal points in the sectors complementary to the sectors where the ellipses are. Consequently, the bisector should be in the sectors where the ellipses are. $\endgroup$ Commented Oct 20, 2021 at 15:01
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Too long for a comment. In an example in Macaulay2

R=QQ[x,y,MonomialOrder=>Lex]

Two ellipses

I=ideal(-3*x^2+2*x*y-2*y^2-17*x+14*y-30,-2*x^2-y^2+14*x+y-18)

And their duals

dI=ideal((71*y^2)/4-59*x*y+25*y+11*x^2-20*x+5,(-13*y^2)+7*x*y+2*y+(71*x^2)/4+14*x+2)

matrix {{131274351*y^4-96177240*y^3-254150568*y^2-116827680*y-15474064, 4357738320*x+65593417383*y^3-77753736324*y^2-93563362092*y-17649191264}}

R=QQ[y,x,MonomialOrder=>Lex]

matrix {{131274351*x^4-57576960*x^3-86987952*x^2+15882240*x+1105856, 11193976176*y-37500706269*x^3-4585249998*x^2+14300319336*x+3683892688}}

So we have the solutions $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ and the relations $131274351y^4-96177240y^3-254150568y^2-116827680y-15474064=0$ that the $y_i$ satisfy and also $131274351x^4-57576960x^3-86987952x^2+15882240x+1105856=0$ that the $x_i$ satisfy

then

$(x x_1+y y_1+1)(x x_2+y y_2+1)(x x_3+y y_3+1)(x x_4+y y_4+1)=0$

or using

ideal((x*x1+y*y1+1)*(x*x2+y*y2+1)*(x*x3+y*y3+1)*(x*x4+y*y4+1),131274351*y4^4-96177240*y4^3-254150568*y4^2-116827680*y4-15474064, 4357738320*x1+65593417383*y1^3-77753736324*y1^2-93563362092*y1-17649191264,131274351*y1^4-96177240*y1^3-254150568*y1^2-116827680*y1-15474064, 4357738320*x2+65593417383*y2^3-77753736324*y2^2-93563362092*y2-17649191264,131274351*y2^4-96177240*y2^3-254150568*y2^2-116827680*y2-15474064, 4357738320*x3+65593417383*y3^3-77753736324*y3^2-93563362092*y3-17649191264,131274351*y3^4-96177240*y3^3-254150568*y3^2-116827680*y3-15474064, 4357738320*x4+65593417383*y4^3-77753736324*y4^2-93563362092*y4-17649191264,131274351*x1+131274351*x2+131274351*x3+131274351*x4-57576960,-131274351*x1*x2-131274351*x1*x3-131274351*x2*x3-131274351*x1*x4-131274351*x2*x4-131274351*x3*x4-86987952,131274351*x1*x2*x3+131274351*x1*x2*x4+131274351*x1*x3*x4+131274351*x2*x3*x4+15882240,-131274351*x1*x2*x3*x4+1105856,(131274351*y1+131274351*y2+131274351*y3+131274351*y4-96177240),(-131274351*y1*y2-131274351*y1*y3-131274351*y2*y3-131274351*y1*y4-131274351*y2*y4-131274351*y3*y4-254150568),(131274351*y1*y2*y3+131274351*y1*y2*y4+131274351*y1*y3*y4+131274351*y2*y3*y4-116827680),-131274351*y1*y2*y3*y4-15474064)

the four tangents are

$1105856x^4+5396992x^3y-5326656x^2y^2-31177472xy^3-15474064y^4-15882240x^3+31083840x^2y+174435840xy^2+116827680y^3-86987952x^2-234816192xy-254150568y^2+57576960x+96177240y+131274351=0.$

If you can find the correct factorization into quadratics, you can hope to use the formula $\tan \theta = {2 \sqrt{h^2-ab} \over a+b}, $ where $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents your line pair.

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  • $\begingroup$ is it possible to express this logic in terms of some algorithm? I am not planning to use that software for this purpose. Let's say I got the four homogenous equations, now is it possible to use the formula you listed here? $\tan \theta = {2 \sqrt{h^2-ab} \over a+b}$ $\endgroup$ Commented Feb 14, 2021 at 23:38
  • $\begingroup$ @KashishDhal: It's possible given two conics, to get the "four" tangents as a completely reducible (over ${\Bbb C}$) equation like the one above i.t.o. the coefficients. If they are ellipses that intersect (over ${\Bbb R}$), the inner tangents are complex. Maybe you can use this to identify the factor you need to use the latter formula. $\endgroup$ Commented Feb 15, 2021 at 7:29

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