Projective Geometry/Classic/Projective Transformations/Transformations of the projective plane

Two-dimensional projective transformations are a type of automorphism of the projective plane onto itself.

Planar transformations can be defined synthetically as follows: point X on a "subjective" plane must be transformed to a point T also on the subjective plane. The transformations uses these tools: a pair of "observation points" P and Q, and an "objective" plane. The subjective and objective planes and the two points all lie in three-dimensional space, and the two planes can intersect at some line.

Draw line l1 through points P and X. Line l1 intersects the objective plane at point R. Draw line l2 through points Q and R. Line l2 intersects the projective plane at point T. Then T is the projective transform of X.

Analysis
Let the xy-plane be the "subjective" plane and let plane m be the "objective" plane. Let plane m be described by


 * $$ z = f(x,y) = m x + n y + b $$

where the constants m and n are partial slopes and b is the z-intercept.

Let there be a pair of "observation" points P and Q,


 * $$ P : (P_x, P_y, P_z), $$
 * $$ Q : (Q_x, Q_y, Q_z). $$

Let point X lie on the "subjective" plane:


 * $$ X : (x,y,0). $$

Point X must be transformed to a point T,


 * $$ T : (T_x, T_y, 0) $$

also on the "subjective" plane.

The analytical results are a pair of equations, one for abscissa Tx and one for ordinate Ty:


 * $$ T_x = {x (-m Q_x P_z - n Q_z P_y + Q_z (P_z - b)) + (n y + b) (Q_z P_x - Q_x P_z) \over (m x + n y) (Q_z - P_z) - (m P_x + n P_y) Q_z + (Q_z - b) P_z}, \qquad \qquad (12) $$


 * $$ T_y = {y (-n Q_y P_z - m Q_z P_x + Q_z (P_z - b)) + (m x + b) (Q_z P_y - Q_y P_z) \over (n y + m x) (Q_z - P_z) - (n P_y + m P_x) Q_z + (Q_z - b) P_z }. \qquad \qquad (13) $$

There are (at most) nine degrees of freedom for defining a 2D transformation: Px, Py, Pz, Qx, Qy, Qz, m, n, b. Notice that equations (12) and (13) have the same denominators, and that Ty can be obtained from Tx by exchanging m with n, and x with y (including subscripts of P and Q).

Trilinear fractional transformations
Let


 * $$ \alpha = -m Q_x P_z - n Q_z P_y + Q_z (P_z - b), $$
 * $$ \beta = n (Q_z P_x - Q_x P_z), $$
 * $$ \gamma = b (Q_z P_x - Q_x P_z), $$
 * $$ \delta = m (Q_z - P_z), $$
 * $$ \epsilon = n (Q_z - P_z), $$
 * $$ \zeta = - (m P_x + n P_y) Q_z + (Q_z - b) P_z, $$

so that


 * $$ T_x = {\alpha x + \beta y + \gamma \over \delta x + \epsilon y + \zeta}. \qquad \qquad (14) $$

Also let


 * $$ \eta = m (Q_z P_y - Q_y P_z), $$
 * $$ \theta = -m Q_z P_x - n Q_y P_z + Q_z (P_z - b), $$
 * $$ \kappa = b (Q_z P_y - Q_y P_z), $$

so that


 * $$ T_y = {\eta x + \theta y + \kappa \over \delta x + \epsilon y + \zeta}. \qquad \qquad (15) $$

Equations (14) and (15) together describe the trilinear fractional transformation.

Composition of trilinear transformations
If a transformation is given by equations (14) and (15), then such transformation is characterized by nine coefficients which can be arranged into a coefficient matrix


 * $$ M_T = \begin{bmatrix} \alpha & \beta & \gamma

\\ \eta & \theta & \kappa \\ \delta & \epsilon & \zeta \end{bmatrix}. $$

If there are a pair T1 and T2 of planar transformations whose coefficient matrices are $$ M_{T_1} $$ and $$ M_{T_2} $$, then the composition of these transformations is another planar transformation T3,


 * $$ T_3 = T_2 \circ T_1, $$

such that


 * $$ T_3(x,y) = T_2 ( T_1 (x,y) ). $$

The coefficient matrix of T3 can be obtained by multiplying the coefficient matrices of T2 and T1:


 * $$ M_{T_3} = M_{T_2} \, M_{T_1}. $$

Proof
Given T1 defined by


 * $$ T_{1x} = {\alpha_1 x + \beta_1 y + \gamma_1 \over \delta_1 x + \epsilon_1 y + \zeta_1}, $$
 * $$ T_{1y} = {\eta_1 x + \theta_1 y + \kappa_1 \over \delta_1 x + \epsilon_1 y + \zeta_1}, $$

and given T2 defined by
 * $$ T_{2x} = {\alpha_2 x + \beta_2 y + \gamma_2 \over \delta_2 x + \epsilon_2 y + \zeta_2}, $$
 * $$ T_{2y} = {\eta_2 x + \theta_2 y + \kappa_2 \over \delta_2 x + \epsilon_2 y + \zeta_2}, $$

then T3 can be calculated by substituting T1 into T2,


 * $$ T_{3x} = T_{2x} ( T_{1x}, T_{1y} ) = { \alpha_2 \left( {\alpha_1 x + \beta_1 y + \gamma_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \beta_2 \left( {\eta_1 x + \theta_1 y + \kappa_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \gamma_2 \over \delta_2 \left( {\alpha_1 x + \beta_1 y + \gamma_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \epsilon_2 \left( {\eta_1 x + \theta_1 y + \kappa_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \zeta_2}. $$

Multiply numerator and denominator by the same trinomial,


 * $$ T_{3x} = {\alpha_2 (\alpha_1 x + \beta_1 y + \gamma_1) + \beta_2 (\eta_1 x + \theta_1 y + \kappa_1) + \gamma_2 (\delta_1 x + \epsilon_1 y + \zeta_1) \over \delta_2 (\alpha_1 x + \beta_1 y + \gamma_1) + \epsilon_2 (\eta_1 x + \theta_1 y + \kappa_1) + \zeta_2 (\delta_1 x + \epsilon_1 y + \zeta_1)}. $$

Group the coefficients of x, y, and 1:


 * $$ T_{3x} = { x (\alpha_2 \alpha_1 + \beta_2 \eta_1 + \gamma_2 \delta_1) + y (\alpha_2 \beta_1 + \beta_2 \theta_1 + \gamma_2 \epsilon_1) + (\alpha_2 \gamma_1 + \beta_2 \kappa_1 + \gamma_2 \zeta_1) \over x (\delta_2 \alpha_1 + \epsilon_2 \eta_1 + \zeta_2 \delta_1) + y (\delta_2 \beta_1 + \epsilon_2 \theta_1 + \zeta_2 \epsilon_1) + (\delta_2 \gamma_1 + \epsilon_2 \kappa_1 + \zeta_2 \zeta_1)} = {\alpha_3 x + \beta_3 y + \gamma_3 \over \delta_3 x + \epsilon_3 y + \zeta_3}. $$

These six coefficients of T3 are the same as those obtained through the product


 * $$ \begin{bmatrix} \alpha_2 & \beta_2 & \gamma_2 \\

\eta_2 & \theta_2 & \kappa_2 \\ \delta_2 & \epsilon_2 & \zeta_2 \end{bmatrix} \begin{bmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \eta_1 & \theta_1 & \kappa_1 \\ \delta_1 & \epsilon_1 & \zeta_1 \end{bmatrix} = \begin{bmatrix} \alpha_3 & \beta_3 & \gamma_3 \\ \eta_3 & \theta_3 & \kappa_3 \\ \delta_3 & \epsilon_3 & \zeta_3 \end{bmatrix}. \qquad \qquad (16) $$

The remaining three coefficients can be verified thus


 * $$ T_{3y} = T_{2y} ( T_{1x}, T_{1y} ) = { \eta_2 \left( {\alpha_1 x + \beta_1 y + \gamma_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \theta_2 \left( {\eta_1 x + \theta_1 y + \kappa_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \kappa_2 \over \delta_2 \left( {\alpha_1 x + \beta_1 y + \gamma_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \epsilon_2 \left( {\eta_1 x + \theta_1 y + \kappa_1 \over \delta_1 x + \epsilon_1 y + \zeta_1} \right) + \zeta_2}. $$

Multiply numerator and denominator by the same trinomial,


 * $$ T_{3y} = {\eta_2 (\alpha_1 x + \beta_1 y + \gamma_1) + \theta_2 (\eta_1 x + \theta_1 y + \kappa_1) + \kappa_2 (\delta_1 x + \epsilon_1 y + \zeta_1) \over \delta_2 (\alpha_1 x + \beta_1 y + \gamma_1) + \epsilon_2 (\eta_1 x + \theta_1 y + \kappa_1) + \zeta_2 (\delta_1 x + \epsilon_1 y + \zeta_1)}. $$

Group the coefficients of x, y, and 1:


 * $$ T_{3x} = { x (\eta_2 \alpha_1 + \theta_2 \eta_1 + \kappa_2 \delta_1) + y (\eta_2 \beta_1 + \theta_2 \theta_1 + \kappa_2 \epsilon_1) + (\eta_2 \gamma_1 + \theta_2 \kappa_1 + \kappa_2 \zeta_1) \over

x (\delta_2 \alpha_1 + \epsilon_2 \eta_1 + \zeta_2 \delta_1) + y (\delta_2 \beta_1 + \epsilon_2 \theta_1 + \zeta_2 \epsilon_1) + (\delta_2 \gamma_1 + \epsilon_2 \kappa_1 + \zeta_2 \zeta_1)} = {\eta_3 x + \theta_3 y + \kappa_3 \over \delta_3 x + \epsilon_3 y + \zeta_3}. $$

The three remaining coefficients just obtained are the same as those obtained through equation (16). Q.E.D.

Planar transformations of lines
The trilinear transformation given be equations (14) and (15) transforms a straight line


 * $$ y = m x + b $$

into another straight line


 * $$ T_y = n T_x + c $$

where n and c are constants and equal to


 * $$ n = {m (\epsilon \kappa - \zeta \theta) + b (\delta \theta - \epsilon \eta) + (\delta \kappa - \zeta \eta) \over m (\epsilon \gamma - \zeta \beta) + b (\delta \beta - \epsilon \alpha) + (\delta \gamma - \zeta \alpha)} $$

and


 * $$ c = {m (\beta \kappa - \gamma \theta) + b (\alpha \theta - \beta \eta) + (\alpha \kappa - \gamma \eta) \over m (\beta \zeta - \gamma \epsilon) + b (\alpha \epsilon - \beta \delta) + (\alpha \zeta - \gamma \delta) }. $$

Proof
Given y = m x + b, then plugging this into equations (14) and (15) yields


 * $$ T_x = {\alpha x + \beta (m x + b) + \gamma \over \delta x + \epsilon (m x + b) + \zeta} = {(\alpha + \beta m) x + (\beta b + \gamma) \over (\delta + \epsilon m) x + (\epsilon b + \zeta)}, $$

and


 * $$ T_y = {(\eta + \theta m) x + (\theta b + \kappa) \over (\delta + \epsilon m) x + (\epsilon b + \zeta) }. $$

If Ty = n Tx + c and n and c are constants, then


 * $$ {\partial T_y \over \partial x} = n {\partial T_x \over \partial x} $$

so that


 * $$ n = {\partial T_y / \partial x \over \partial T_x / \partial y}. $$

Calculation shows that
 * $$ {\partial T_x \over \partial x} = { (\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m) \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)]^2 } $$

and


 * $$ {\partial T_y \over \partial x} = { (\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m) \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)]^2 } $$

therefore


 * $$ n = {\partial T_y / \partial x \over \partial T_x / \partial y} =

{ (\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m) \over (\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m) }. $$

We should now obtain c to be


 * $$ c = T_y - n T_x $$
 * $$ = {(\eta + \theta m) x + (\theta b + \kappa) - \left[ { (\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m) \over (\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m) } \right] \cdot [ (\alpha + \beta m) x + (\beta b + \gamma) ] \over (\delta + \epsilon m) x + (\epsilon b + \zeta) }. $$

Add the two fractions in the numerator:
 * $$ c = { \left\{ [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] [(\eta + \theta m) x + (\theta b + \kappa)] - [(\epsilon b + \zeta) (\eta + \theta m) - (\theta b + \kappa) (\delta + \epsilon m)] [(\alpha + \beta m) x + (\beta b + \gamma)] \right\}

\over [(\delta + \epsilon m) x + (\epsilon b + \zeta)] [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] }. $$

Distribute binomials in parentheses in the numerator, then cancel out equal and opposite terms:


 * $$ c = { - (\beta b + \gamma) (\delta + \epsilon m) (\eta + \theta m) x + (\epsilon b + \zeta) (\alpha + \beta m) (\theta b + \kappa) + (\theta b + \kappa) (\delta + \epsilon m) (\alpha + \beta m) x - (\epsilon b + \zeta) (\eta + \theta m) (\beta b + \gamma) \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)] [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] }. $$

Factor the numerator into a pair of terms, only one of them having the numerus cossicus (x). There is another numerus cossicus in the denominator. The objective now is to get both of these to cancel out.


 * $$ c = { \left\{ [(\theta b + \kappa) (\alpha + \beta m) - (\beta b + \gamma) (\eta + \theta m)] (\delta + \epsilon m) x + [(\alpha + \beta m)(\theta b + \kappa) - (\eta + \theta m) (\beta b + \gamma)] (\epsilon b + \zeta) \right\} \over [(\delta + \epsilon m) x + (\epsilon b + \zeta)] [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] }. $$

Factor the numerator,


 * $$ c = {[(\theta b + \kappa) (\alpha + \beta m) - (\beta b + \gamma) (\eta + \theta m)] [(\delta + \epsilon m) x + (\epsilon b + \zeta)] \over [(\epsilon b + \zeta) (\alpha + \beta m) - (\beta b + \gamma) (\delta + \epsilon m)] [(\delta + \epsilon m) x + (\epsilon b + \zeta)] }. $$

The terms with the numeri cossici cancel out, therefore


 * $$ c = { (\alpha + \beta m) (\theta b + \kappa) - (\beta b + \gamma) (\eta + \theta m) \over (\alpha + \beta m) (\epsilon b + \zeta) - (\beta b + \gamma) (\delta + \epsilon m) } $$

is a constant. Q.E.D.

Comparing c with n, notice that their denominators are the same. Also, n is obtained from c by exchanging the following coefficients:


 * $$ \alpha \leftrightarrow \delta, \ \beta \leftrightarrow \epsilon, \ \gamma \leftrightarrow \zeta . $$

There is also the following exchange symmetry between the numerator and denominator of n:


 * $$ \alpha \leftrightarrow \eta, \ \beta \leftrightarrow \theta, \ \gamma \leftrightarrow \kappa . $$

The numerator and denominator of c also have exchange symmetry: $$ \{ \eta \leftrightarrow \delta, \ \theta \leftrightarrow \epsilon, \ \kappa \leftrightarrow \zeta \}. $$

The exchange symmetry between n and c can be chunked into binomials:


 * $$ n \leftrightarrow c \equiv \{ (\alpha + m \beta ) \leftrightarrow (\delta + m \epsilon ), \ (\gamma + b \beta ) \leftrightarrow (\zeta + b \epsilon ) \}. $$

All of these exchange symmetries amount to exchanging pairs of rows in the coefficient matrix.

Planar transformations of conic sections
A trilinear transformation such as T given by equations (14) and (15) will convert a conic section
 * $$ A x^2 + B y^2 + C x + D y + E x y + F = 0 \qquad \qquad (17) $$

into another conic section
 * $$ A' T_x^2 + B' T_y^2 + C' T_x + D' T_y + E' T_x T_y + F' = 0. \qquad \qquad (18) $$

Proof
Let there be given a conic section described by equation (17) and a planar transformation T described by equations (15) and (16) which converts points (x,y) into points (Tx,Ty).

It is possible to find an inverse transformation T′ which converts back points (Tx,Ty) to points (x,y). This inverse transformation has a coefficient matrix
 * $$ M_{T'} = \begin{bmatrix} \alpha' & \beta' & \gamma' \\

\eta' & \theta' & \kappa' \\ \delta' & \epsilon' & \zeta' \end{bmatrix}. $$

Equation (17) can be expressed in terms of the inverse transformation:


 * $$ A \left( {\alpha' T_x + \beta' T_y + \gamma' \over \delta' T_x + \epsilon' T_y + \zeta'} \right)^2 + B \left( {\eta' T_x + \theta' T_y + \kappa' \over \delta' T_x + \epsilon' T_y + \zeta'} \right)^2 + C \left( {\alpha' T_x + \beta' T_y + \gamma' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) + D \left( {\eta' T_x + \theta' T_y + \kappa' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) + E \left( {\alpha' T_x + \beta' T_y + \gamma' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) \left( {\eta' T_x + \theta' T_y + \kappa' \over \delta' T_x + \epsilon' T_y + \zeta'} \right) + F = 0. $$

The denominators can be "dissolved" by multiplying both sides of the equation by the square of a trinomial:


 * $$ A (\alpha' T_x + \beta' T_y + \gamma')^2 + B (\eta' T_x + \theta' T_y + \kappa')^2 + C (\alpha' T_x + \beta' T_y + \gamma') (\delta' T_x + \epsilon' T_y + \zeta') + D (\eta' T_x + \theta' T_y + \kappa') (\delta' T_x + \epsilon' T_y + \zeta') + E (\alpha' T_x + \beta' T_y + \gamma') (\eta' T_x + \theta' T_y + \kappa') + F (\delta' T_x + \epsilon' T_y + \zeta')^2 = 0. $$

Expand the products of trinomials and collect common powers of Tx and Ty:


 * $$ \begin{matrix}

(A \alpha'^2 + B \eta'^2 + C \alpha' \delta' + D \eta' \delta' + E \alpha' \eta' + F \delta'^2) T_x^2 \\ + (A \beta'^2 + B \theta'^2 + C \beta' \epsilon' + D \theta' \epsilon' + E \beta' \theta' + F \epsilon'^2) T_y^2 \\ + (2 A \alpha' \gamma' + 2 B \eta' \kappa' + C (\alpha' \zeta' + \gamma' \delta') + D (\eta' \zeta' + \kappa' \delta') + E (\alpha' \kappa' + \gamma' \eta') + 2 F \delta' \zeta') T_x \\ + (2 A \beta' \gamma' + 2 B \theta' \kappa' + C (\beta' \zeta' + \gamma' \epsilon') + D (\theta' \zeta' + \kappa' \epsilon') + E (\beta' \kappa' + \gamma' \theta') + 2 F \epsilon' \zeta') T_y \\ + (2 A \alpha' \beta' + 2 B \eta' \theta' + C (\alpha' \epsilon' + \beta' \delta') + D (\eta' \epsilon' + \theta' \delta') + E (\alpha' \theta' + \beta' \eta') + 2 F \delta' \epsilon') T_x T_y \\ + (A \gamma'^2 + B \kappa'^2 + C \gamma' \zeta' + D \kappa' \zeta' + E \gamma' \kappa' + F \zeta'^2) = 0. \end{matrix} \qquad \qquad (19) $$

Equation (19) has the same form as equation (18).

What remains to do is to express the primed coefficients in terms of the unprimed coefficients. To do this, apply Cramer's rule to the coefficient matrix MT to obtain the primed matrix of the inverse transformation:


 * $$ M_{T'} = {1 \over \Delta} \begin{bmatrix}

\left| \begin{matrix} \theta &\kappa \\ \epsilon & \zeta \end{matrix} \right| & \left| \begin{matrix} \epsilon & \zeta \\ \beta & \gamma \end{matrix} \right| & \left| \begin{matrix} \beta & \gamma \\ \theta & \kappa \end{matrix} \right| \\ \quad & \quad & \quad \\ \left| \begin{matrix} \kappa & \eta \\ \zeta & \delta \end{matrix} \right| & \left| \begin{matrix} \zeta & \delta \\ \gamma & \alpha \end{matrix} \right| & \left| \begin{matrix} \gamma & \alpha \\ \kappa & \eta \end{matrix} \right| \\ \quad & \quad & \quad \\ \left| \begin{matrix} \eta & \theta \\ \delta & \epsilon \end{matrix} \right| & \left| \begin{matrix} \delta &\epsilon \\ \alpha & \beta \end{matrix} \right| & \left| \begin{matrix} \alpha & \beta \\ \eta & \theta \end{matrix} \right| \end{bmatrix} \qquad \qquad (20) $$ where Δ is the determinant of the unprimed coefficient matrix.

Equation (20) allows primed coefficients to be expressed in terms of unprimed coefficients. But performing these substitutions on the primed coefficients of equation (19) it can be noticed that the determinant Δ cancels itself out, so that it can be ignored altogether. Therefore
 * $$ A' = A (\theta \zeta - \kappa \epsilon)^2

+ B (\kappa \delta - \eta \zeta)^2 + C (\theta \zeta - \kappa \epsilon) (\eta \epsilon - \theta \delta) + D (\kappa \delta - \eta \zeta) (\eta \epsilon - \theta \delta) + E (\theta \zeta - \kappa \epsilon) (\kappa \delta - \eta \zeta) + F (\eta \epsilon - \theta \delta)^2 $$


 * $$ B' = A (\epsilon \gamma - \zeta \beta)^2

+ B (\zeta \alpha - \delta \gamma)^2 + C (\epsilon \gamma - \zeta \beta) (\delta \beta - \epsilon \alpha) + D (\zeta \alpha - \delta \gamma) (\delta \beta - \epsilon \alpha) + E (\epsilon \gamma - \zeta \beta) (\zeta \alpha - \delta \gamma) + F (\delta \beta - \epsilon \alpha)^2 $$


 * $$ C' = 2 A (\theta \zeta - \kappa \epsilon) (\beta \kappa - \gamma \theta)

+ 2 B (\kappa \delta - \eta \zeta) (\gamma \eta - \alpha \kappa) + C [ (\theta \zeta - \kappa \epsilon) (\alpha \theta - \beta \eta) + (\beta \kappa - \gamma \theta) (\eta \epsilon - \theta \delta)] + D [ (\kappa \delta - \eta \zeta) (\alpha \theta - \beta \eta) + (\gamma \eta - \alpha \kappa) (\eta \epsilon - \theta \delta) ] + E [ (\theta \zeta - \kappa \epsilon) (\gamma \eta - \alpha \kappa) + (\beta \kappa - \gamma \theta) (\kappa \delta - \eta \zeta) ] + 2 F (\eta \epsilon - \theta \delta) (\alpha \theta - \beta \eta) $$


 * $$ D' = 2 A (\epsilon \gamma - \zeta \beta) (\beta \kappa - \gamma \theta)

+ 2 B (\zeta \alpha - \delta \gamma) (\gamma \eta - \alpha \kappa) + C [ (\epsilon \gamma - \zeta \beta) (\alpha \theta - \beta \eta) + (\beta \kappa - \gamma \theta) (\delta \beta - \epsilon \alpha) ] + D [ (\zeta \alpha - \delta \gamma) (\alpha \theta - \beta \eta) + (\gamma \eta - \alpha \kappa) (\delta \beta - \epsilon \alpha) ] + E [ (\epsilon \gamma - \zeta \beta) (\gamma \eta - \alpha \kappa) + (\beta \kappa - \gamma \theta) (\zeta \alpha - \delta \gamma) ] + 2 F (\delta \beta - \epsilon \alpha) (\alpha \theta - \beta \eta) $$


 * $$ E' = 2 A (\theta \zeta - \kappa \epsilon) (\epsilon \gamma - \zeta \beta)

+ 2 B (\kappa \delta - \eta \zeta) (\zeta \alpha - \delta \gamma) + C [(\theta \zeta - \kappa \epsilon) (\delta \beta - \epsilon \alpha) + (\epsilon \gamma - \zeta \beta) (\eta \epsilon - \theta \delta)] + D [ (\kappa \delta - \eta \zeta) (\delta \beta - \epsilon \alpha) + (\zeta \alpha - \delta \gamma) (\eta \epsilon - \theta \delta)] + E [ (\theta \zeta - \kappa \epsilon) (\zeta \alpha - \delta \gamma) + (\epsilon \gamma - \zeta \beta) (\kappa \delta - \eta \zeta)] + 2 F (\eta \epsilon - \theta \delta) (\delta \beta - \epsilon \alpha) $$


 * $$ F' = A (\beta \kappa - \gamma \theta)^2

+ B (\gamma \eta - \alpha \kappa)^2 + C (\beta \kappa - \gamma \theta) (\alpha \theta - \beta \eta) + D (\gamma \eta - \alpha \kappa) (\alpha \theta - \beta \eta) + E (\beta \kappa - \gamma \theta) (\gamma \eta - \alpha \kappa) + F (\alpha \theta - \beta \eta)^2 $$

The coefficients of the transformed conic have been expressed in terms of the coefficients of the original conic and the coefficients of the planar transformation T. Q.E.D.

Planar projectivities and cross-ratio
Let four points A, B, C, D be collinear. Let there be a planar projectivity T which transforms these points into points A′, B′, C′, and D′. It was already shown that lines are transformed into lines, so that the transformed points A′ through D′ will also be collinear. Then it will turn out that the cross-ratio of the original four points is the same as the cross-ratio of their transforms:
 * $$ [A \ B \ C \ D] = [A' \ B' \ C' \ D']. $$

Proof
If the two-dimensional coordinates of four points are known, and if the four points are collinear, then their cross-ratio can be found from their abscissas alone. It is possible to project the points onto a horizontal line by means of a pencil of vertical lines issuing from a point on the line at infinity:
 * $$ [A \ B \ C \ D] = [A_x \ B_x \ C_x \ D_x]. $$

The same is true for the ordinates of the points. The reason is that any mere rescaling of the coordinates of the points does not change the cross-ratio.

Let
 * $$ A : (x_1, m x_1 + b), $$
 * $$ B : (x_2, m x_2 + b), $$
 * $$ C : (x_3, m x_3 + b), $$
 * $$ D : (x_4, m x_4 + b). $$

Clearly these four points are collinear. Let


 * $$ T_x (x,y) = {\alpha x + \beta y + \gamma \over \delta x + \epsilon y + \zeta} $$

be the first half of a trilinear transformation. Then


 * $$ T_x(A) = {\alpha x_1 + \beta (m x_1 + b) + \gamma \over \delta x_1 + \epsilon (m x_1 + b) + \zeta} = {(\alpha + \beta m) x_1 + (\beta b + \gamma) \over (\delta + \epsilon m) x_1 + (\epsilon b + \zeta)}, $$
 * $$ T_x(B) = {\alpha x_2 + \beta (m x_2 + b) + \gamma \over \delta x_2 + \epsilon (m x_2 + b) + \zeta} = {(\alpha + \beta m) x_2 + (\beta b + \gamma) \over (\delta + \epsilon m) x_2 + (\epsilon b + \zeta)}, $$
 * $$ T_x(C) = {\alpha x_3 + \beta (m x_3 + b) + \gamma \over \delta x_3 + \epsilon (m x_3 + b) + \zeta} = {(\alpha + \beta m) x_3 + (\beta b + \gamma) \over (\delta + \epsilon m) x_3 + (\epsilon b + \zeta)}, $$
 * $$ T_x(D) = {\alpha x_4 + \beta (m x_4 + b) + \gamma \over \delta x_4 + \epsilon (m x_4 + b) + \zeta} = {(\alpha + \beta m) x_4 + (\beta b + \gamma) \over (\delta + \epsilon m) x_4 + (\epsilon b + \zeta)}. $$

The original cross-ratio is


 * $$ [x_1 \ x_2 \ x_3 \ x_4] = {x_1 - x_3 \over x_1 - x_4} \cdot {x_2 - x_4 \over x_2 - x_3}. $$

It is not necessary to calculate the transformed cross-ratio. Just let


 * $$ S(x) = {(\alpha + \beta m) x + (\beta b + \gamma) \over (\delta + \epsilon m) x + (\epsilon b + \zeta)} $$

be a bilinear transformation. Then S(x) is a one-dimensional projective transformation. But Tx(A)=S(A), Tx(B)=S(B), Tx(C)=S(C), and Tx(D)=S(D). Therefore


 * $$ [T_x(A) \ T_x(B) \ T_x(C) \ T_x(D)] = [S(A) \ S(B) \ S(C) \ S(D)] $$

but it has already been shown that bilinear fractional transformations preserve cross-ratio. Q.E.D.

Example
The following is a rather simple example of a planar projectivity:
 * $$ T_x = {1 \over x}, \qquad T_y = {y \over x}. $$

The coefficient matrix of this projectivity T is
 * $$ M_T = \begin{bmatrix} 0 & 0 & 1 \\

0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} $$ and it is easy to verify that MT is its own inverse.

The locus of points described parametrically as $$ ( \cos \theta, \, \sin \theta ) $$ describe a circle, due to the trigonometric identity
 * $$ \cos^2 \theta + \sin^2 \theta = 1 $$

which has the same form as the canonical equation of a circle. Applying the projectivity T yields the locus of points described parametrically by $$ (\sec \theta,\, \tan \theta) $$ which describe a hyperbola, due to the trigonometric identity
 * $$ \sec^2 \theta - \tan^2 \theta = 1 $$

which has the same form as the canonical equation of a hyperbola. Notice that points $$ (~-1,0) $$ and $$(1,0)$$ are fixed points.

Indeed, this projectivity transforms any circle, of any radius, into a hyperbola centered at the origin with both of its foci lying on the x-axis, and vice versa. This projectivity also transforms the y-axis into the line at infinity, and vice versa:
 * $$ T : (0, y) \rightarrow \left( {1 \over 0}, {y \over 0} \right) = (\pm \infty, \pm \infty), $$
 * $$ T: (\pm \infty, \pm \infty) \rightarrow \left( {1 \over \pm \infty}, {\pm \infty \over \pm \infty} \right) = (0, y). $$

The ratio of infinity over infinity is indeterminate which means that it can be set to any value y desired.

This example emphasizes that in the real projective plane, RP², a hyperbola is a closed curve which passes twice through the line at infinity. But what does the transformation do to a parabola?

Let the locus of points $$ (x,x^2) $$ describe a parabola. Its transformation is
 * $$ T : (x,x^2) \rightarrow \left( {1 \over x}, {x^2 \over x} \right) = (x', 1/x') $$

which is a hyperbola whose asymptotes are the x-axis and the y-axis and whose wings lie in the first quadrant and the third quadrant. Likewise, the hyperbola
 * $$ y = {1 \over x} $$

is transformed by T into the parabola
 * $$ y = x^2 \quad $$.

On the other hand, the parabola described by the locus of points $$ (x, \pm \sqrt{x}) $$ is transformed by T into itself: this demonstrates that a parabola intersects the line at infinity at a single point.