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Orthogonal trajectories of curves

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hello !
I have to find the orthogonal trajectories of the curves :

[tex] x^{2}-y^{2}=c , x^{2}+y^{2}+2cy=1 [/tex] ..

How can I do this??

For this: [tex] x^{2}-y^{2}=c [/tex] I found [tex] \left | y \right |=\frac{M}{\left | x \right |} [/tex] ,and for this: [tex] x^{2}+y^{2}+2cy=1 [/tex],I found: [tex]y=Ax-D[/tex],c,A,D [tex] \varepsilon \Re [/tex] .

Is this right??And how can I continue??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hello !
I have to find the orthogonal trajectories of the curves :

[tex] x^{2}-y^{2}=c , x^{2}+y^{2}+2cy=1 [/tex] ..

How can I do this??

For this: [tex] x^{2}-y^{2}=c [/tex] I found [tex] \left | y \right |=\frac{M}{\left | x \right |} [/tex] ,and for this: [tex] x^{2}+y^{2}+2cy=1 [/tex],I found: [tex]y=Ax-D[/tex],c,A,D [tex] \varepsilon \Re [/tex] .

Is this right??And how can I continue??
Hi evinda! :D

That is right enough.
Actually, you can sharpen the second one a bit since $D=c$.

What do you want to continue with?
Aren't you already done?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hi evinda! :D

That is right enough.
Actually, you can sharpen the second one a bit since $D=c$.

What do you want to continue with?
Aren't you already done?
Since both of the curves are given at the same subquestion,don't I have to find something common between those two solutions?? :confused: Or,what do I have to do?? :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Since both of the curves are given at the same subquestion,don't I have to find something common between those two solutions?? :confused: Or,what do I have to do?? :confused:
The solution for the first set of curves is a set of hyperboles.
The solution for the second set of curves is a set of lines.
They have nothing in common.
So in my opinion they are separate questions and your answer is sufficient.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
The solution for the first set of curves is a set of hyperboles.
The solution for the second set of curves is a set of lines.
They have nothing in common.
So in my opinion they are separate questions and your answer is sufficient.
Nice..Thank you very much! :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hold on! :eek:

Something was still bugging me about that $c$ in the second set of curves $$x^2+y^2+2cy=1 \qquad (1)$$
When I started to draw them, I realized that
$$y=Ax-c \qquad \qquad (2)$$
is not the proper solution.
Draw them. You'll see!


Those lines are only orthogonal for points on 1 specific circle that is defined by $c$.
To find the curves that are orthogonal to the whole set of circles, we need to eliminate $c$.

From (1), we get that:
\begin{array}{}
\frac{x^2}{y}+y+2c&=&\frac 1 y \\
\frac{2x}{y}dx - \frac{x^2}{y^2}dy + dy + 0 &=& -\frac{1}{y^2}dy \\
2xydx&=&(x^2-y^2-1)dy
\end{array}
See, no $c$ involved anymore! ;)
The solution of this differential equation is the set of curves (actually circles with their center on the y axis) in (1).
x2+y2+2cy=1.png

To find the orthogonal curves, we need to solve the "orthogonal" differential equation:
\begin{array}{}
(x^2-y^2-1)dx&=&-2xydy \\
\Big(1 - \frac{y^2}{x^2} - \frac 1 {x^2} \Big)dx &=& - \frac{2y}{x}dy \\
dx + \left(- \frac{y^2}{x^2}dx + \frac{2y}{x}dy \right) - \frac 1 {x^2}dx &=& 0 \\
x + \frac{y^2}{x} + \frac{1}{x} &=& 2C \\
x^2 + y^2 - 2Cx &=& -1 \\
(x - C)^2 + y^2 &=& C^2 - 1
\end{array}
In other words, this is a set of circles (with their center on the x-axis) instead of a set of lines.
orthogonal_x2+y2+2cy=1.png
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hold on! :eek:

Something was still bugging me about that $c$ in the second set of curves $$x^2+y^2+2cy=1 \qquad (1)$$
When I started to draw them, I realized that
$$y=Ax-c \qquad \qquad (2)$$
is not the proper solution.
Draw them. You'll see!


Those lines are only orthogonal for points on 1 specific circle that is defined by $c$.
To find the curves that are orthogonal to the whole set of circles, we need to eliminate $c$.

From (1), we get that:
\begin{array}{}
\frac{x^2}{y}+y+2c&=&\frac 1 y \\
\frac{2x}{y}dx - \frac{x^2}{y^2}dy + dy + 0 &=& -\frac{1}{y^2}dy \\
2xydx&=&(x^2-y^2-1)dy
\end{array}
See, no $c$ involved anymore! ;)
The solution of this differential equation is the set of curves (actually circles with their center on the y axis) in (1).
View attachment 1702

To find the orthogonal curves, we need to solve the "orthogonal" differential equation:
\begin{array}{}
(x^2-y^2-1)dx&=&-2xydy \\
\Big(1 - \frac{y^2}{x^2} - \frac 1 {x^2} \Big)dx &=& - \frac{2y}{x}dy \\
dx + \left(- \frac{y^2}{x^2}dx + \frac{2y}{x}dy \right) - \frac 1 {x^2}dx &=& 0 \\
x + \frac{y^2}{x} + \frac{1}{x} &=& 2C \\
x^2 + y^2 - 2Cx &=& -1 \\
(x - C)^2 + y^2 &=& C^2 - 1
\end{array}
In other words, this is a set of circles (with their center on the x-axis) instead of a set of lines.
View attachment 1703
I understood it now :eek: Thank you very much!!! :)