How to make mirrored parabola
I try to figure it out how to make mirror of the parabola to some line.
For example like that:
In that example my original parabola is: $f_p(x) = x^2 $ - (red line)
My mirror center line is: $f_c(x) = x $ - (green line)
And my mirrored parabola is: $f_m(x) = sqrt { x } $ - (blue line - of course it's just half of parabola but it's enough for me)
But what in case if my mirror center line is for example: $f_c(x) = 0.3x$
And my parabola that I want to mirror is more complicated like: $f_p(x) = ax^2+bx+c$
How to make mirrored parabola to such line (just for remind: half of parabola would be great enough for me).
For any help thanks in advance.
geometry conic-sections
asked Nov 27 '18 at 22:32
pajczur
854
add a comment |
I try to figure it out how to make mirror of the parabola to some line.
For example like that:
In that example my original parabola is: $f_p(x) = x^2 $ - (red line)
My mirror center line is: $f_c(x) = x $ - (green line)
And my mirrored parabola is: $f_m(x) = sqrt { x } $ - (blue line - of course it's just half of parabola but it's enough for me)
But what in case if my mirror center line is for example: $f_c(x) = 0.3x$
And my parabola that I want to mirror is more complicated like: $f_p(x) = ax^2+bx+c$
How to make mirrored parabola to such line (just for remind: half of parabola would be great enough for me).
For any help thanks in advance.
geometry conic-sections
asked Nov 27 '18 at 22:32
pajczur
854
1
Notice that $x mapsto sqrt{x}$ may locally look like the mirrored parabola, but it's actually not. The mirrored parabola is given by $y^2 = x$.
– Viktor Glombik
Nov 27 '18 at 22:42
1
The only reflections that will result in a function of $x$ are those in vertical lines. For reflections in arbitrary lines, you’ll need to use an implicit equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.
– amd
Nov 27 '18 at 23:07
add a comment |
I try to figure it out how to make mirror of the parabola to some line.
For example like that:
In that example my original parabola is: $f_p(x) = x^2 $ - (red line)
My mirror center line is: $f_c(x) = x $ - (green line)
And my mirrored parabola is: $f_m(x) = sqrt { x } $ - (blue line - of course it's just half of parabola but it's enough for me)
But what in case if my mirror center line is for example: $f_c(x) = 0.3x$
And my parabola that I want to mirror is more complicated like: $f_p(x) = ax^2+bx+c$
How to make mirrored parabola to such line (just for remind: half of parabola would be great enough for me).
For any help thanks in advance.
geometry conic-sections
asked Nov 27 '18 at 22:32
pajczur
854
I try to figure it out how to make mirror of the parabola to some line.
For example like that:
In that example my original parabola is: $f_p(x) = x^2 $ - (red line)
My mirror center line is: $f_c(x) = x $ - (green line)
And my mirrored parabola is: $f_m(x) = sqrt { x } $ - (blue line - of course it's just half of parabola but it's enough for me)
But what in case if my mirror center line is for example: $f_c(x) = 0.3x$
And my parabola that I want to mirror is more complicated like: $f_p(x) = ax^2+bx+c$
How to make mirrored parabola to such line (just for remind: half of parabola would be great enough for me).
For any help thanks in advance.
geometry conic-sections
geometry conic-sections
asked Nov 27 '18 at 22:32
pajczur
854
asked Nov 27 '18 at 22:32
pajczur
854
asked Nov 27 '18 at 22:32
pajczur
854
asked Nov 27 '18 at 22:32
pajczur
854
asked Nov 27 '18 at 22:32
pajczur
854
854
1
Notice that $x mapsto sqrt{x}$ may locally look like the mirrored parabola, but it's actually not. The mirrored parabola is given by $y^2 = x$.
– Viktor Glombik
Nov 27 '18 at 22:42
1
The only reflections that will result in a function of $x$ are those in vertical lines. For reflections in arbitrary lines, you’ll need to use an implicit equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.
– amd
Nov 27 '18 at 23:07
add a comment |
1
Notice that $x mapsto sqrt{x}$ may locally look like the mirrored parabola, but it's actually not. The mirrored parabola is given by $y^2 = x$.
– Viktor Glombik
Nov 27 '18 at 22:42
1
The only reflections that will result in a function of $x$ are those in vertical lines. For reflections in arbitrary lines, you’ll need to use an implicit equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.
– amd
Nov 27 '18 at 23:07
1
1
Notice that $x mapsto sqrt{x}$ may locally look like the mirrored parabola, but it's actually not. The mirrored parabola is given by $y^2 = x$.
– Viktor Glombik
Nov 27 '18 at 22:42
Notice that $x mapsto sqrt{x}$ may locally look like the mirrored parabola, but it's actually not. The mirrored parabola is given by $y^2 = x$.
– Viktor Glombik
Nov 27 '18 at 22:42
1
1
The only reflections that will result in a function of $x$ are those in vertical lines. For reflections in arbitrary lines, you’ll need to use an implicit equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.
– amd
Nov 27 '18 at 23:07
The only reflections that will result in a function of $x$ are those in vertical lines. For reflections in arbitrary lines, you’ll need to use an implicit equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.
– amd
Nov 27 '18 at 23:07
add a comment |
2 Answers
2
active
oldest
votes
For simple parabolas of the form $f(x) = ax^2$ the mirrored (about $y = x$) parabola is given by $y^2 = frac{x}{a}$.
The intuition here is to swap the variables to achieve the mirroring:
begin{equation*}
f(x) = y = ax^2 iff frac{y}{a} = x^2
overset{textrm{mirror}}{implies} y^2 = frac{x}{a}.
end{equation*}
The same goes for parabolas of the form $(x-a)^2$ or $(x-a)(x-b)$, just swap the variables ($x = y$ and vice versa, since you reflect upon $y = x$!).
This covered all parabolas.
Note that an equation of the form $y^2 = cx$ is not a function of x since there are two $y$-values for every $x$-value. If you only want the positive part (analogous to the root-function) you can solve the above equation for one of these $x$-values.
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
2
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
add a comment |
Point $P=(x,y)$ is reflected about line $y=mx$ to a point $P'=(x',y')$ given by:
$$
x'={2my+(1-m^2)xover1+m^2},quad y'={2mx-(1-m^2)yover1+m^2}.
$$
To obtain the reflected equation of parabola $y=x^2$ it is then enough to substitute the above expressions into $y'=x'^2$, to get:
$$
4m^2y^2+4m(1-m^2)xy+(1-m^2)^2 x^2+(1-m^4)y-2m(1+m^2)x=0.
$$
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
add a comment |
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2 Answers
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2 Answers
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active
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active
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votes
active
oldest
votes
For simple parabolas of the form $f(x) = ax^2$ the mirrored (about $y = x$) parabola is given by $y^2 = frac{x}{a}$.
The intuition here is to swap the variables to achieve the mirroring:
begin{equation*}
f(x) = y = ax^2 iff frac{y}{a} = x^2
overset{textrm{mirror}}{implies} y^2 = frac{x}{a}.
end{equation*}
The same goes for parabolas of the form $(x-a)^2$ or $(x-a)(x-b)$, just swap the variables ($x = y$ and vice versa, since you reflect upon $y = x$!).
This covered all parabolas.
Note that an equation of the form $y^2 = cx$ is not a function of x since there are two $y$-values for every $x$-value. If you only want the positive part (analogous to the root-function) you can solve the above equation for one of these $x$-values.
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
2
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
add a comment |
For simple parabolas of the form $f(x) = ax^2$ the mirrored (about $y = x$) parabola is given by $y^2 = frac{x}{a}$.
The intuition here is to swap the variables to achieve the mirroring:
begin{equation*}
f(x) = y = ax^2 iff frac{y}{a} = x^2
overset{textrm{mirror}}{implies} y^2 = frac{x}{a}.
end{equation*}
The same goes for parabolas of the form $(x-a)^2$ or $(x-a)(x-b)$, just swap the variables ($x = y$ and vice versa, since you reflect upon $y = x$!).
This covered all parabolas.
Note that an equation of the form $y^2 = cx$ is not a function of x since there are two $y$-values for every $x$-value. If you only want the positive part (analogous to the root-function) you can solve the above equation for one of these $x$-values.
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
2
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
add a comment |
For simple parabolas of the form $f(x) = ax^2$ the mirrored (about $y = x$) parabola is given by $y^2 = frac{x}{a}$.
The intuition here is to swap the variables to achieve the mirroring:
begin{equation*}
f(x) = y = ax^2 iff frac{y}{a} = x^2
overset{textrm{mirror}}{implies} y^2 = frac{x}{a}.
end{equation*}
The same goes for parabolas of the form $(x-a)^2$ or $(x-a)(x-b)$, just swap the variables ($x = y$ and vice versa, since you reflect upon $y = x$!).
This covered all parabolas.
Note that an equation of the form $y^2 = cx$ is not a function of x since there are two $y$-values for every $x$-value. If you only want the positive part (analogous to the root-function) you can solve the above equation for one of these $x$-values.
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
For simple parabolas of the form $f(x) = ax^2$ the mirrored (about $y = x$) parabola is given by $y^2 = frac{x}{a}$.
The intuition here is to swap the variables to achieve the mirroring:
begin{equation*}
f(x) = y = ax^2 iff frac{y}{a} = x^2
overset{textrm{mirror}}{implies} y^2 = frac{x}{a}.
end{equation*}
The same goes for parabolas of the form $(x-a)^2$ or $(x-a)(x-b)$, just swap the variables ($x = y$ and vice versa, since you reflect upon $y = x$!).
This covered all parabolas.
Note that an equation of the form $y^2 = cx$ is not a function of x since there are two $y$-values for every $x$-value. If you only want the positive part (analogous to the root-function) you can solve the above equation for one of these $x$-values.
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
edited Nov 27 '18 at 23:51
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
answered Nov 27 '18 at 22:50
Viktor Glombik
576425
576425
2
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
add a comment |
2
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
2
2
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
But it doesn't cover reflecting in other than $y=x,$ which post asks about.
– coffeemath
Nov 27 '18 at 23:06
add a comment |
Point $P=(x,y)$ is reflected about line $y=mx$ to a point $P'=(x',y')$ given by:
$$
x'={2my+(1-m^2)xover1+m^2},quad y'={2mx-(1-m^2)yover1+m^2}.
$$
To obtain the reflected equation of parabola $y=x^2$ it is then enough to substitute the above expressions into $y'=x'^2$, to get:
$$
4m^2y^2+4m(1-m^2)xy+(1-m^2)^2 x^2+(1-m^4)y-2m(1+m^2)x=0.
$$
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
add a comment |
Point $P=(x,y)$ is reflected about line $y=mx$ to a point $P'=(x',y')$ given by:
$$
x'={2my+(1-m^2)xover1+m^2},quad y'={2mx-(1-m^2)yover1+m^2}.
$$
To obtain the reflected equation of parabola $y=x^2$ it is then enough to substitute the above expressions into $y'=x'^2$, to get:
$$
4m^2y^2+4m(1-m^2)xy+(1-m^2)^2 x^2+(1-m^4)y-2m(1+m^2)x=0.
$$
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
add a comment |
Point $P=(x,y)$ is reflected about line $y=mx$ to a point $P'=(x',y')$ given by:
$$
x'={2my+(1-m^2)xover1+m^2},quad y'={2mx-(1-m^2)yover1+m^2}.
$$
To obtain the reflected equation of parabola $y=x^2$ it is then enough to substitute the above expressions into $y'=x'^2$, to get:
$$
4m^2y^2+4m(1-m^2)xy+(1-m^2)^2 x^2+(1-m^4)y-2m(1+m^2)x=0.
$$
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
Point $P=(x,y)$ is reflected about line $y=mx$ to a point $P'=(x',y')$ given by:
$$
x'={2my+(1-m^2)xover1+m^2},quad y'={2mx-(1-m^2)yover1+m^2}.
$$
To obtain the reflected equation of parabola $y=x^2$ it is then enough to substitute the above expressions into $y'=x'^2$, to get:
$$
4m^2y^2+4m(1-m^2)xy+(1-m^2)^2 x^2+(1-m^4)y-2m(1+m^2)x=0.
$$
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
edited Nov 29 '18 at 18:31
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
answered Nov 27 '18 at 23:07
Aretino
22.6k21442
22.6k21442
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
add a comment |
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
And this works also for a parabola of the form $x mapsto ax^2 + bx + c$?
– Viktor Glombik
Nov 27 '18 at 23:45
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
@ViktorGlombik Of course it works in that case too.
– Aretino
Nov 28 '18 at 9:26
add a comment |
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1
Notice that $x mapsto sqrt{x}$ may locally look like the mirrored parabola, but it's actually not. The mirrored parabola is given by $y^2 = x$.
– Viktor Glombik
Nov 27 '18 at 22:42
1
The only reflections that will result in a function of $x$ are those in vertical lines. For reflections in arbitrary lines, you’ll need to use an implicit equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.
– amd
Nov 27 '18 at 23:07