I need to learn how to find the properties (foci, vertex, etc) of a conic given its equation, I know how to do it when the coeficient of x y = 0 , but I have troubles in the other case. Can someone recommend me a book, website o document where I can learn how to find the properties of a conic section given its equation please?

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I need to learn how to find the properties (foci, vertex, etc) of a conic given its equation, I know how to do it when the coeficient of   x  y  =  0 , but I have troubles in the other case. Can someone recommend me a book, website o document where I can learn how to find the properties of a conic section given its equation please?

Gornil2 · Accepted Answer

Step 1Given the equation of the conic in the form  A      x    2    +  B  x  y  +  C      y    2    +  D  x  +  E  y  +  F  =  0The term containing xy can be eliminated using a rotation. Define the change of variables as follows  x  =      x    ′    cos  ⁡  θ  −      y    ′    sin  ⁡  θ  y  =      x    ′    sin  ⁡  θ  +      y    ′    cos  ⁡  θA point (x, y) will have coordinates   (      x    ′    ,      y    ′    ) in a coordinate frame that is rotated by   θ counter-clockwise with respect to the original frame.Plug these in,                          A                  (                      x                          ′              2                                            cos            2                    ⁡          θ          +                      y                          ′              2                                            sin            2                    ⁡          θ          −                      x            ′                                y            ′                    sin          ⁡          2          θ          )                                                  +        B                  (                                    1              2                                (                      x                          ′              2                                −                      y                          ′              2                                )          sin          ⁡          2          θ          +                      x            ′                                y            ′                    cos          ⁡          2          θ          )                                                  +        C                  (                      x                          ′              2                                            sin            2                    ⁡          θ          +                      y                          ′              2                                            cos            2                    ⁡          θ          +                      x            ′                                y            ′                    sin          ⁡          2          θ          )                                                  +        D                  (                      x            ′                    cos          ⁡          θ          −                      y            ′                    sin          ⁡          θ          )                +        E                  (                      x            ′                    sin          ⁡          θ          +                      y            ′                    cos          ⁡          θ          )                +        F        =        0            Combining like terms,                                    x                      ′            2                                    (          A                      cos            2                    ⁡          θ          +          C                      sin            2                    ⁡          θ          +                                    1              2                                B          sin          ⁡          2          θ          )                                                  +                  y                      ′            2                                    (          A                      sin            2                    ⁡          θ          +          C                      cos            2                    ⁡          θ          −                                    1              2                                sin          ⁡          2          θ          )                                                  +                  x          ′                          y          ′                          (          (          C          −          A          )          sin          ⁡          2          θ          +          B          cos          ⁡          2          θ          )                                                  +        D                  (                      x            ′                    cos          ⁡          θ          −                      y            ′                    sin          ⁡          θ          )                +        E                  (                      x            ′                    sin          ⁡          θ          +                      y            ′                    cos          ⁡          θ          )                +        F        =        0            Now we want to eliminate the       x    ′        y    ′   term, so we have to choose   θ such that  (  C  −  A  )  sin  ⁡  2  θ  +  B  cos  ⁡  2  θ  =  0that is,  tan  ⁡  2  θ  =            B              A        −        C            Now the equation is      A    ′        x          ′      2        +      C    ′        y          ′      2        +      D    ′        x    ′    +      E    ′        y    ′    +      F    ′    =  0where      A    ′    =  A      cos    2    ⁡  θ  +  C      sin    2    ⁡  θ  +            1      2        B  sin  ⁡  2  θ      C    ′    =  A      sin    2    ⁡  θ  +  C      cos    2    ⁡  θ  −            1      2        B  sin  ⁡  2  θ      D    ′    =  D  cos  ⁡  θ  +  E  sin  ⁡  θ      E    ′    =  −  D  sin  ⁡  θ  +  E  cos  ⁡  θ      F    ′    =  FSince rotation preserves lengths, you can find the axes, foci, vertices, and then rotate everything by the angle   θ to obtain them in the original coordinate frame.

I need to learn how to find the properties (foci, vertex, etc) of a conic given its equation, I know

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