Drawing bezier curve from a parabola
0
$begingroup$
I'm not a math guy, sorry. I read posts on the subject but couldn't find the answer to my problem (or didn't understood the answers). I'd like to get a simple answer.
I know a generic parabola formula
$$(Pa.x+Pb.y)^2+Pc.x+Pd.y+Pe=0$$
it focus, $(Fx,Fy)$
and the cartesian equation of it's directrix
$$Da.x+Db.y+Dc=0$$
What would be the easiest way to draw a portion of it using a quadratic bezier curve (say maybe from it summit up to $2p$ or like) ?
conic-sections bezier-curve
edited Dec 6 '18 at 20:40
dantopa
6,46942243
asked Dec 6 '18 at 20:26
fdesarfdesar
11
$endgroup$
add a comment |
0
$begingroup$
I'm not a math guy, sorry. I read posts on the subject but couldn't find the answer to my problem (or didn't understood the answers). I'd like to get a simple answer.
I know a generic parabola formula
$$(Pa.x+Pb.y)^2+Pc.x+Pd.y+Pe=0$$
it focus, $(Fx,Fy)$
and the cartesian equation of it's directrix
$$Da.x+Db.y+Dc=0$$
What would be the easiest way to draw a portion of it using a quadratic bezier curve (say maybe from it summit up to $2p$ or like) ?
conic-sections bezier-curve
edited Dec 6 '18 at 20:40
dantopa
6,46942243
asked Dec 6 '18 at 20:26
fdesarfdesar
11
$endgroup$
$begingroup$
Welcome the MathematicsStackExchange! You can improve your experience by using MathJax for equations. A quick review is at math.meta.stackexchange.com/questions/5020/….
$endgroup$
– dantopa
Dec 6 '18 at 20:39
$begingroup$
How are you “drawing” the curve? What do you consider easy?
$endgroup$
– amd
Dec 6 '18 at 21:06
$begingroup$
I want to draw the curve using SVG path Q/q command. Want I mean by "easy" is with the less computation possible.
$endgroup$
– fdesar
Dec 7 '18 at 22:18
$begingroup$
See here: en.wikipedia.org/wiki/…
$endgroup$
– Aretino
Dec 8 '18 at 21:50
$begingroup$
Do you know how to compute the position and first derivative at any point on the parabola from your "generic parabola formula"?
$endgroup$
– fang
Dec 8 '18 at 22:55
add a comment |
0
0
0
$begingroup$
I'm not a math guy, sorry. I read posts on the subject but couldn't find the answer to my problem (or didn't understood the answers). I'd like to get a simple answer.
I know a generic parabola formula
$$(Pa.x+Pb.y)^2+Pc.x+Pd.y+Pe=0$$
it focus, $(Fx,Fy)$
and the cartesian equation of it's directrix
$$Da.x+Db.y+Dc=0$$
What would be the easiest way to draw a portion of it using a quadratic bezier curve (say maybe from it summit up to $2p$ or like) ?
conic-sections bezier-curve
edited Dec 6 '18 at 20:40
dantopa
6,46942243
asked Dec 6 '18 at 20:26
fdesarfdesar
11
$endgroup$
I'm not a math guy, sorry. I read posts on the subject but couldn't find the answer to my problem (or didn't understood the answers). I'd like to get a simple answer.
I know a generic parabola formula
$$(Pa.x+Pb.y)^2+Pc.x+Pd.y+Pe=0$$
it focus, $(Fx,Fy)$
and the cartesian equation of it's directrix
$$Da.x+Db.y+Dc=0$$
What would be the easiest way to draw a portion of it using a quadratic bezier curve (say maybe from it summit up to $2p$ or like) ?
conic-sections bezier-curve
conic-sections bezier-curve
edited Dec 6 '18 at 20:40
dantopa
6,46942243
asked Dec 6 '18 at 20:26
fdesarfdesar
11
edited Dec 6 '18 at 20:40
dantopa
6,46942243
asked Dec 6 '18 at 20:26
fdesarfdesar
11
edited Dec 6 '18 at 20:40
dantopa
6,46942243
edited Dec 6 '18 at 20:40
dantopa
6,46942243
edited Dec 6 '18 at 20:40
dantopa
6,46942243
6,46942243
asked Dec 6 '18 at 20:26
fdesarfdesar
11
asked Dec 6 '18 at 20:26
fdesarfdesar
11
asked Dec 6 '18 at 20:26
fdesarfdesar
11
11
$begingroup$
Welcome the MathematicsStackExchange! You can improve your experience by using MathJax for equations. A quick review is at math.meta.stackexchange.com/questions/5020/….
$endgroup$
– dantopa
Dec 6 '18 at 20:39
$begingroup$
How are you “drawing” the curve? What do you consider easy?
$endgroup$
– amd
Dec 6 '18 at 21:06
$begingroup$
I want to draw the curve using SVG path Q/q command. Want I mean by "easy" is with the less computation possible.
$endgroup$
– fdesar
Dec 7 '18 at 22:18
$begingroup$
See here: en.wikipedia.org/wiki/…
$endgroup$
– Aretino
Dec 8 '18 at 21:50
$begingroup$
Do you know how to compute the position and first derivative at any point on the parabola from your "generic parabola formula"?
$endgroup$
– fang
Dec 8 '18 at 22:55
add a comment |
$begingroup$
Welcome the MathematicsStackExchange! You can improve your experience by using MathJax for equations. A quick review is at math.meta.stackexchange.com/questions/5020/….
$endgroup$
– dantopa
Dec 6 '18 at 20:39
$begingroup$
How are you “drawing” the curve? What do you consider easy?
$endgroup$
– amd
Dec 6 '18 at 21:06
$begingroup$
I want to draw the curve using SVG path Q/q command. Want I mean by "easy" is with the less computation possible.
$endgroup$
– fdesar
Dec 7 '18 at 22:18
$begingroup$
See here: en.wikipedia.org/wiki/…
$endgroup$
– Aretino
Dec 8 '18 at 21:50
$begingroup$
Do you know how to compute the position and first derivative at any point on the parabola from your "generic parabola formula"?
$endgroup$
– fang
Dec 8 '18 at 22:55
$begingroup$
Welcome the MathematicsStackExchange! You can improve your experience by using MathJax for equations. A quick review is at math.meta.stackexchange.com/questions/5020/….
$endgroup$
– dantopa
Dec 6 '18 at 20:39
$begingroup$
Welcome the MathematicsStackExchange! You can improve your experience by using MathJax for equations. A quick review is at math.meta.stackexchange.com/questions/5020/….
$endgroup$
– dantopa
Dec 6 '18 at 20:39
$begingroup$
How are you “drawing” the curve? What do you consider easy?
$endgroup$
– amd
Dec 6 '18 at 21:06
$begingroup$
How are you “drawing” the curve? What do you consider easy?
$endgroup$
– amd
Dec 6 '18 at 21:06
$begingroup$
I want to draw the curve using SVG path Q/q command. Want I mean by "easy" is with the less computation possible.
$endgroup$
– fdesar
Dec 7 '18 at 22:18
$begingroup$
I want to draw the curve using SVG path Q/q command. Want I mean by "easy" is with the less computation possible.
$endgroup$
– fdesar
Dec 7 '18 at 22:18
$begingroup$
See here: en.wikipedia.org/wiki/…
$endgroup$
– Aretino
Dec 8 '18 at 21:50
$begingroup$
See here: en.wikipedia.org/wiki/…
$endgroup$
– Aretino
Dec 8 '18 at 21:50
$begingroup$
Do you know how to compute the position and first derivative at any point on the parabola from your "generic parabola formula"?
$endgroup$
– fang
Dec 8 '18 at 22:55
$begingroup$
Do you know how to compute the position and first derivative at any point on the parabola from your "generic parabola formula"?
$endgroup$
– fang
Dec 8 '18 at 22:55
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Choose at will start point $P_0$ and end point $P_2$ on your parabola.
Find the equations of tangent lines at $P_0$ and $P_2$.
Find the intersection point $P_1$ of the two tangents.
The quadratic Bézier $B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2$ is what you want.
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
$endgroup$
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Choose at will start point $P_0$ and end point $P_2$ on your parabola.
Find the equations of tangent lines at $P_0$ and $P_2$.
Find the intersection point $P_1$ of the two tangents.
The quadratic Bézier $B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2$ is what you want.
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
$endgroup$
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
add a comment |
0
$begingroup$
Choose at will start point $P_0$ and end point $P_2$ on your parabola.
Find the equations of tangent lines at $P_0$ and $P_2$.
Find the intersection point $P_1$ of the two tangents.
The quadratic Bézier $B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2$ is what you want.
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
$endgroup$
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
add a comment |
0
0
0
$begingroup$
Choose at will start point $P_0$ and end point $P_2$ on your parabola.
Find the equations of tangent lines at $P_0$ and $P_2$.
Find the intersection point $P_1$ of the two tangents.
The quadratic Bézier $B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2$ is what you want.
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
$endgroup$
Choose at will start point $P_0$ and end point $P_2$ on your parabola.
Find the equations of tangent lines at $P_0$ and $P_2$.
Find the intersection point $P_1$ of the two tangents.
The quadratic Bézier $B(t)=(1-t)^2P_0+2t(1-t)P_1+t^2P_2$ is what you want.
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
answered Dec 9 '18 at 9:42
AretinoAretino
23.1k21443
23.1k21443
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
add a comment |
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
Souds logical. My problem is to choose p0 and p2.
$endgroup$
– fdesar
Dec 9 '18 at 23:41
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
I interpret "from its summit" as meaning that $P_0$ is the vertex of the parabola. And that's easy because the vertex is the midpoint between the focus and the projection of the focus on the directrix.The meaning of "up to $2p$ or like" is otherwise rather obscure, to me at least.
$endgroup$
– Aretino
Dec 10 '18 at 11:58
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Yep F is the focus and p is the parabola parameter, which is the length of the orthogonal projection of F on the directrix.
$endgroup$
– fdesar
Dec 11 '18 at 22:05
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Do you then want to choose $P_2$ such that its distance from the directrix is $2p$? If so, $P_2$ is the intersection between the parabola and one of the two lines with equation $D_a x+D_b y+D_cpm 2psqrt{D_a^2+D_b^2}=0$.
$endgroup$
– Aretino
Dec 12 '18 at 18:30
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
$begingroup$
Thank you. Sounds logical.
$endgroup$
– fdesar
Dec 14 '18 at 8:05
add a comment |
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$begingroup$
Welcome the MathematicsStackExchange! You can improve your experience by using MathJax for equations. A quick review is at math.meta.stackexchange.com/questions/5020/….
$endgroup$
– dantopa
Dec 6 '18 at 20:39
$begingroup$
How are you “drawing” the curve? What do you consider easy?
$endgroup$
– amd
Dec 6 '18 at 21:06
$begingroup$
I want to draw the curve using SVG path Q/q command. Want I mean by "easy" is with the less computation possible.
$endgroup$
– fdesar
Dec 7 '18 at 22:18
$begingroup$
See here: en.wikipedia.org/wiki/…
$endgroup$
– Aretino
Dec 8 '18 at 21:50
$begingroup$
Do you know how to compute the position and first derivative at any point on the parabola from your "generic parabola formula"?
$endgroup$
– fang
Dec 8 '18 at 22:55