Point(s) of Intersection Between 2 Conics

1. Let's consider 2 Conics represented by General Quadratic Equations in 2 Variables as follows
   
   \(A_1x^2 + B_1xy + C_1y^2 + D_1x + E_1y + F_1=0\)   ...(1)
   
   \(A_2x^2 + B_2xy + C_2y^2 + D_2x + E_2y + F_2=0\)   ...(2)
   
   The 2 Conics given by equations (1) and (2) above can be realted in following ways
   1. The 2 Conics may be Coincident.
   2. The 2 Conics may be Distinct (i.e Not Intersect at all).
   3. The 2 Conics may have From 1 to 4 Real or Imaginary Points of Intersection. If All Points of Intersections are Imaginary, it is same as Not Having Any Point of Intersection at all on a Real Plane.
2. The following gives the Steps to Calculate the Point of Intersection(s) Betweeb 2 Conics as give by equations (1) and (2) above
   
   If the Co-efficient of \(x^2\) is Non-Zero in either equation (1) or (2) or Both then do the following steps
   1. If \(A_1 \neq 0\) and \(A_2 \neq 0\) then divide equations (1) and (2) by \(A_1\) and \(A_2\) repectively as follows
      
      \(x^2 + {\Large \frac{B_1}{A_1}}xy + {\Large \frac{C_1}{A_1}}y^2 + {\Large \frac{D_1}{A_1}}x + {\Large \frac{E_1}{A_1}}y + {\Large \frac{F_1}{A_1}}=0\)   ...(3)
      
      \(x^2 + {\Large \frac{B_2}{A_2}}xy + {\Large \frac{C_2}{A_2}}y^2 + {\Large \frac{D_2}{A_2}}x + {\Large \frac{E_2}{A_2}}y + {\Large \frac{F_2}{A_2}}=0\)   ...(4)
      
      Subtract equation (3) from equation (4) to calculte the Difference Equation without the \(x^2\) term as follows
      
      \((A_2B_1 - A_1B_2) xy + (A_2C_1 - A_1C_2)y^2 + (A_2D_1 - A_1D_2)x + (A_2E_1 - A_1E_2)y + (A_2F_1 - A_1F_2)=0\)   ...(5)
      
      Setting \(B' = A_2B_1 - A_1B_2\), \(C' = A_2C_1 - A_1C_2\), \(D' = A_2D_1 - A_1D_2\), \(E' = A_2E_1 - A_1E_2\) and \(F' = A_2F_1 - A_1F_2\) we get
      
      \(B'xy + C'y^2 + D'x + E'y + F'=0\)   ...(6)
      
      If \(A_1 = 0\) then swap equations (1) and (2) so that \(A_1\) becomes Non Zero and \(A_2\) becomes Zero (i.e. equation (2) becomes the equation without the \(x^2\) term)
      
      Then Setting \(B' = B_2\), \(C' = C_2\), \(D' = D_2\), \(E' = E_2\) and \(F' = F_2\) in equation (2) we get
      
      \(B'xy + C'y^2 + D'x + E'y + F'=0\)   ...(7)
   2. If in equations (6) or (7) (i.e. equations without the \(x^2\) term) \(B'=C'=D'=E'=F'=0\), then Both the Conics given by equations (1) and (2) are Coincident.
   3. If in equations (6) or (7) (i.e. equations without the \(x^2\) term) \(B'=C'=D'=E'=0\) and \(F'\neq 0\), then Both the Conics given by equations (1) and (2) are Distinct and have No Points of Intersection.
   4. If in equations (6) or (7) (i.e. equations without the \(x^2\) term) \(B'=D'=0\), then these become either Quadratic Equations in \(y\) (if \(C'\neq 0\)) or Linear Equations in \(y\) (if \(C'= 0\)) from where the value(s) of variable \(y\) can be found out. For each value of \(y\) the corresponding value of \(x\) can be calculated by putting the value of \(y\) in equation (1) (i.e. one with the Non Zero Value of \(x^2\) term). These value of \(x\) and \(y\) give the Point(s) of Intersection of the 2 Conics given by equation (1) and (2).
   5. If in equations (6) or (7) (i.e. equations without the \(x^2\) term) either \(B'\neq 0\) or \(D'\neq 0\), then find the value of \(x\) as follows
      
      \(x={\Large \frac{C'y^2 + E'y + F}{-B'y - D'}}\)   ...(8)
      
      Substitute the value of \(x\) obtained through equation (7) in equation (1) (i.e. one with the Non Zero Value of \(x^2\) term) as follows
      
      \(A_1{\Large (\frac{C'y^2 + E'y + F'}{-B'y - D'})}^2 + B_1y{\Large \frac{C'y^2 + E'y + F}{-B'y - D'}} + C_1y^2 + D_1{\Large \frac{C'y^2 + E'y + F}{-B'y - D'}} + E_1y + F_1=0\)
      
      \(\Rightarrow A_1(C'y^2 + E'y + F')^2 - (B_1y + D_1)(C'y^2 + E'y + F)(B'y + D') + (C_1y^2 + E_1y + F_1){(B'y + D')}^2=0\)   ...(9)
      
      \(\Rightarrow (C_1{B'}^2 + A_1{C'}^2-B_1B'C')y^4 + (2C_1B'D'+E_1{B'}^2+2A_1C'E'-B_1B'E'-B_1D'C'-D_1B'C')y^3 + \\ (F_1{B'}^2 + 2E_1B'D' + C_1{D'}^2 + A_1{E'}^2 +2A_1C'F' - B_1B'F' + B_1D'E' -D_1B'E'-D_1D'C')y^2 + \\ (2F_1B'D' + 2A_1E'F' + E_1{D'}2 -B_1D'F' - D_1B'F' - D_1D'E')y + (F_1{D'}^2 + A_1{F'}^2 -D_1D'F')=0\)   ...(10)
      
      Equation (10) above gives a Quartic or Cubic or Quadratic Equation in terms of \(y\) which can be solved to get the value(s) of variable \(y\). For each value of \(y\) the corresponding value of \(x\) can be calculated by putting the value of \(y\) in equation (8) above. These value of \(x\) and \(y\) give the Point(s) of Intersection of the 2 Conics given by equation (1) and (2).
   If the Co-efficient of \(x^2\) is Zero in Both equations (1) and (2), but the Co-efficient of \(y^2\) is Non-Zero in either equation (1) or (2) or Both then do the following steps
   1. If \(C_1 \neq 0\) and \(C_2 \neq 0\) then divide equations (1) and (2) by \(C_1\) and \(C_2\) repectively as follows
      
      \({\Large \frac{A_1}{C_1}}x^2 + {\Large \frac{B_1}{C_1}}xy + y^2 + {\Large \frac{D_1}{C_1}}x + {\Large \frac{E_1}{C_1}}y + {\Large \frac{F_1}{C_1}}=0\)   ...(11)
      
      \({\Large \frac{A_2}{C_2}}x^2 + {\Large \frac{B_2}{C_2}}xy + y^2 + {\Large \frac{D_2}{C_2}}x + {\Large \frac{E_2}{C_2}}y + {\Large \frac{F_2}{C_2}}=0\)   ...(12)
      
      Subtract equation (12) from equation (11) to calculte the Difference Equation without the \(y^2\) term as follows
      
      \((C_2A_1 - C_1A_2)x^2 + (C_2B_1 - C_1B_2) xy + (C_2D_1 - C_1D_2)x + (C_2E_1 - C_1E_2)y + (C_2F_1 - C_1F_2)=0\)   ...(13)
      
      Setting \(A' = C_2A_1 - C_1A_2\), \(B' = C_2B_1 - C_1B_2\), \(D' = C_2D_1 - C_1D_2\), \(E' = C_2E_1 - C_1E_2\) and \(F' = C_2F_1 - C_1F_2\) we get
      
      \(A'x^2 + B'xy + D'x + E'y + F'=0\)   ...(14)
      
      If \(C_1 = 0\) then swap equations (1) and (2) so that \(C_1\) becomes Non Zero and \(C_2\) becomes Zero (i.e. equation (2) becomes the equation without the \(y^2\) term)
      
      Then Setting \(A' = A_2\), \(B' = B_2\), \(D' = D_2\), \(E' = E_2\) and \(F' = F_2\) in equation (2) we get
      
      \(A'x^2 + B'xy + D'x + E'y + F'=0\)   ...(15)
   2. If in equations (14) or (15) (i.e. equations without the \(y^2\) term) \(A'=B'=D'=E'=F'=0\), then Both the Conics given by equations (1) and (2) are Coincident.
   3. If in equations (14) or (15) (i.e. equations without the \(y^2\) term) \(A'=B'=D'=E'=0\) and \(F'\neq 0\), then Both the Conics given by equations (1) and (2) are Distinct and have No Points of Intersection.
   4. If in equations (14) or (15) (i.e. equations without the \(y^2\) term) \(B'=E'=0\), then these become either Quadratic Equations in \(x\) (if \(A'\neq 0\)) or Linear Equations in \(x\) (if \(A'= 0\)) from where the value(s) of variable \(x\) can be found out. For each value of \(x\) the corresponding value of \(y\) can be calculated by putting the value of \(x\) in equation (1) (i.e. one with the Non Zero Value of \(y^2\) term). These value of \(x\) and \(y\) give the Point(s) of Intersection of the 2 Conics given by equation (1) and (2).
   5. If in equations (14) or (15) (i.e. equations without the \(y^2\) term) either \(B'\neq 0\) or \(E'\neq 0\), then find the value of \(y\) as follows
      
      \(y={\Large \frac{A'x^2 + D'x + F'}{-B'x - E'}}\)   ...(16)
      
      Substitute the value of \(y\) obtained through equation (16) in equation (1) (i.e. one with the Non Zero Value of \(y^2\) term) as follows
      
      \(A_1x^2 + B_1x{\Large \frac{A'x^2 + D'x + F'}{-B'x - E'}} + C_1{\Large (\frac{A'x^2 + D'x + F'}{-B'x - E'})}^2 + D_1x + E_1{\Large \frac{A'x^2 + D'x + F'}{-B'x - E'}} + F_1=0\)
      
      \(\Rightarrow (A_1x^2 + D_1x + + F_1){(B'x + E')}^2 - (B_1x + E_1) (A'x^2 + D'x + F')(B'x + E') + C_1{(A'x^2 + D'x + F')}^2 =0\)   ...(17)
      
      \(\Rightarrow (A_1{B'}^2 + C_1{A'}^2 + B_1B'A')x^4 + (2A_1B'E' + 2C_1A'D' + D_1{B'}^2 - E_1A'B' -B_1A'E' - B_1B'D')x^3 + \\ (F_1{B'}^2 + 2D_1B'E' +A_1{E'}^2 + C_1{D'}^2 + 2C_1A'F' -E_1A'E' - E_1B'D' -B_1D'E' - B_1B'F')x^2 + \\ (2F_1B'E' + D_1{E'}^2 + 2C_1D'F'-E_1D'E' - E_1B'F' - B_1E'F')x + (F_1{E'}^2 + C_1{F'}^2 - E_1E'F')=0\)   ...(18)
      
      Equation (18) above gives a Quartic or Cubic or Quadratic Equation in terms of \(x\) which can be solved to get the value(s) of variable \(x\). For each value of \(x\) the corresponding value of \(y\) can be calculated by putting the value of \(x\) in equation (16) above. These value of \(x\) and \(y\) give the Point(s) of Intersection of the 2 Conics given by equation (1) and (2).
   If the Co-efficient of \(x^2\) and \(y^2\) Both are Zero in Both equations (1) and (2), and the Co-efficient of \(xy\) is Non-Zero in both equations (1) and (2) then do the following steps
   1. Divide equations (1) and (2) by \(B_1\) and \(B_2\) repectively as follows
      
      \(xy + {\Large \frac{D_1}{B_1}}x + {\Large \frac{E_1}{B_1}}y + {\Large \frac{F_1}{B_1}}=0\)   ...(19)
      
      \(xy + {\Large \frac{D_2}{B_2}}x + {\Large \frac{E_2}{B_2}}y + {\Large \frac{F_2}{B_2}}=0\)   ...(20)
      
      Subtract equation (20) from equation (19) to calculte the Difference Equation without the \(xy\) term as follows
      
      \((B_2D_1 - B_1D_2)x + (B_2E_1 - B_1E_2)y + (B_2F_1 - B_1F_2)=0\)   ...(21)
      
      Setting \(D' = B_2D_1 - B_1D_2\), \(E' = B_2E_1 - B_1E_2\) and \(F' = B_2F_1 - B_1F_2\) we get
      
      \(D'x + E'y + F'=0\)   ...(22)
   2. If in equation (22) \(D'=E'=F'=0\), then Both the Conics given by equations (1) and (2) are Coincident.
   3. If in equation (22) \(D'=E'=0\) and \(F'\neq 0\), then Both the Conics given by equations (1) and (2) are Distinct and have No Points of Intersection.
   4. If in equation (22) \(E'\neq0\) then calculate the value of \(y\) as
      
      \(y={\Large \frac{-D'x - F'}{E'}}\)   ...(23)
      
      Substitute the value of \(y\) obtained through equation (23) in equation (1) (i.e. one with the Zero Values of \(x^2\) and \(y^2\) terms and Non Zero Value of \(xy\) term) as follows
      
      \(B_1x{\Large \frac{-D'x - F'}{E'}} + D_1x + E_1{\Large \frac{-D'x - F'}{E'}} + F_1=0\)
      
      \(\Rightarrow B_1D'x^2 + (B_1F' + E_1D' - E'D_1)x + E_1F' - E'F_1=0\)   ...(24)
      
      The equation (24) above is either a Quadratic Equation in \(x\) (if \(D'\neq 0\)) or Linear Equation in \(x\) (if \(D'= 0\)) from where the value(s) of variable \(x\) can be found out. For each value of \(x\) the corresponding value of \(y\) can be calculated by putting the value of \(x\) in equation (23) above. These value of \(x\) and \(y\) give the Point(s) of Intersection of the 2 Conics given by equation (1) and (2).
   5. If in equation (22) \(D'\neq0\) then calculate the value of \(x\) as
      
      \(x={\Large \frac{-E'y - F'}{D'}}\)   ...(25)
      
      Substitute the value of \(x\) obtained through equation (25) in equation (1) (i.e. one with the Zero Values of \(x^2\) and \(y^2\) terms and Non Zero Value of \(xy\) term) as follows
      
      \(B_1y{\Large \frac{-E'y - F'}{D'}} + D_1{\Large \frac{-E'y - F'}{D'}} + E_1y + F_1=0\)
      
      \(\Rightarrow B_1E'y^2 + (B_1F' + D_1E' - D'E_1)y + D_1F' - D'F_1=0\)   ...(26)
      
      The equation (26) above is either a Quadratic Equation in \(y\) (if \(E'\neq 0\)) or Linear Equation in \(y\) (if \(E'= 0\)) from where the value(s) of variable \(y\) can be found out. For each value of \(y\) the corresponding value of \(x\) can be calculated by putting the value of \(y\) in equation (25) above. These value of \(x\) and \(y\) give the Point(s) of Intersection of the 2 Conics given by equation (1) and (2).