In differential geometry and geometric optics, a caustic is the envelope of rays either reflected or refracted by a manifold. It is related to the concept of caustics in optics. The ray's source may be a point (called the radiant) or parallel rays from a point at infinity, in which case a direction vector of the rays must be specified.
More generally, especially as applied to symplectic geometry and singularity theory, a caustic is the critical value set of a Lagrangian mapping (π ○ i) : L ↪ M ↠ B; where i : L ↪ M is a Lagrangian immersion of a Lagrangian submanifold L into a symplectic manifold M, and π : M ↠ B is a Lagrangian fibration of the symplectic manifold M. The caustic is a subset of the Lagrangian fibration's base space B.
A catacaustic is the reflective case.
With a radiant, it is the evolute of the orthotomic of the radiant.
The planar, parallel-source-rays case: suppose the direction vector is                     (        a        ,        b        )                 and the mirror curve is parametrised as                     (        u        (        t        )        ,        v        (        t        )        )                . The normal vector at a point is                     (        −                  v          ′                (        t        )        ,                  u          ′                (        t        )        )                ; the reflection of the direction vector is (normal needs special normalization)
                    2                              proj                                n                          d        −        d        =                                            2              n                                      n              ⋅              n                                                                          n              ⋅              d                                      n              ⋅              n                                      −        d        =        2        n                                            n              ⋅              d                                      n              ⋅              n                                      −        d        =                                            (              a                              v                                  ′                                      2                                                              −              2              b                              u                ′                                            v                ′                            −              a                              u                                  ′                                      2                                                              ,              b                              u                                  ′                                      2                                                              −              2              a                              u                ′                                            v                ′                            −              b                              v                                  ′                                      2                                                              )                                                      v                                  ′                                      2                                                              +                              u                                  ′                                      2                                                                                              Having components of found reflected vector treat it as a tangent
                    (        x        −        u        )        (        b                  u                      ′                          2                                      −        2        a                  u          ′                          v          ′                −        b                  v                      ′                          2                                      )        =        (        y        −        v        )        (        a                  v                      ′                          2                                      −        2        b                  u          ′                          v          ′                −        a                  u                      ′                          2                                      )        .                Using the simplest envelope form
                    F        (        x        ,        y        ,        t        )        =        (        x        −        u        )        (        b                  u                      ′                          2                                      −        2        a                  u          ′                          v          ′                −        b                  v                      ′                          2                                      )        −        (        y        −        v        )        (        a                  v                      ′                          2                                      −        2        b                  u          ′                          v          ′                −        a                  u                      ′                          2                                      )                                     =        x        (        b                  u                      ′                          2                                      −        2        a                  u          ′                          v          ′                −        b                  v                      ′                          2                                      )        −        y        (        a                  v                      ′                          2                                      −        2        b                  u          ′                          v          ′                −        a                  u                      ′                          2                                      )        +        b        (        u                  v                      ′                          2                                      −        u                  u                      ′                          2                                      −        2        v                  u          ′                          v          ′                )        +        a        (        −        v                  u                      ′                          2                                      +        v                  v                      ′                          2                                      +        2        u                  u          ′                          v          ′                )                                              F                      t                          (        x        ,        y        ,        t        )        =        2        x        (        b                  u          ′                          u          ″                −        a        (                  u          ′                          v          ″                +                  u          ″                          v          ′                )        −        b                  v          ′                          v          ″                )        −        2        y        (        a                  v          ′                          v          ″                −        b        (                  u          ″                          v          ′                +                  u          ′                          v          ″                )        −        a                  u          ′                          u          ″                )        +        b        (                  u          ′                          v                      ′                          2                                      +        2        u                  v          ′                          v          ″                −                  u                      ′                          3                                      −        2        u                  u          ′                          u          ″                −        2                  u          ′                          v                      ′                          2                                      −        2                  u          ″                v                  v          ′                −        2                  u          ′                v                  v          ″                )        +        a        (        −                  v          ′                          u                      ′                          2                                      −        2        v                  u          ′                          u          ″                +                  v                      ′                          3                                      +        2        v                  v          ′                          v          ″                +        2                  v          ′                          u                      ′                          2                                      +        2                  v          ″                u                  u          ′                +        2                  v          ′                u                  u          ″                )                which may be unaesthetic, but                     F        =                  F                      t                          =        0                 gives a linear system in                     (        x        ,        y        )                 and so it is elementary to obtain a parametrisation of the catacaustic. Cramer's rule would serve.
Let the direction vector be (0,1) and the mirror be                     (        t        ,                  t                      2                          )        .                 Then
                              u          ′                =        1                                                 u          ″                =        0                                                 v          ′                =        2        t                                                 v          ″                =        2                                       a        =        0                                       b        =        1                                    F        (        x        ,        y        ,        t        )        =        (        x        −        t        )        (        1        −        4                  t                      2                          )        +        4        t        (        y        −                  t                      2                          )        =        x        (        1        −        4                  t                      2                          )        +        4        t        y        −        t                                              F                      t                          (        x        ,        y        ,        t        )        =        −        8        t        x        +        4        y        −        1                and                     F        =                  F                      t                          =        0                 has solution                     (        0        ,        1                  /                4        )                ; i.e., light entering a parabolic mirror parallel to its axis is reflected through the focus.