Symmetric bezier curve with unsymmetric control points











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I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
$$
C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
$$

where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.



If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?



I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.










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    up vote
    1
    down vote

    favorite












    I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
    $$
    C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
    $$

    where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.



    If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?



    I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.










    share|cite|improve this question









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    Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
      $$
      C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
      $$

      where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.



      If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?



      I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.










      share|cite|improve this question









      New contributor




      Donguk Jung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm solving this problem about Cubic Bezier Curve which have $4$ control points $C_0,C_1,C_2,C_3$.
      $$
      C_0 = (0,0),: C_1 = (x_1,y_1),: C_2=(x_2,y_2),: C_3=(1,0)
      $$

      where $0 < x_1 < x_2 < 1$ and $y_1, y_2>0$.



      If $C_1$ and $C_2$ are symmetric with respect to axis $x = 0.5$ (such as $C_1 = (0.25,1)$, $C_2 = (0.75,1)$), obviously whole bezier curve is symmetric (w.r.t. $x=0.5$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $C_1,C_2$?



      I tried with reduction, if $C_1, C_2$ is not symmetric and the image $B(C_0,C_1,C_2,C_3)$ is symmetric, there should be $C_1^prime$ that symmetric with respect to $C_1$ but $C_1^prime := C2 cdot B(C_0,C_1,C_1^prime,C_3)$ is symmetric and when I move just one control point $C_1^prime$ to $C2$, I'm not sure that symmetric property will broke or not.







      bezier-curve






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      edited yesterday









      Coolwater

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      asked 2 days ago









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