Orthogonal trajectories of curves

evinda

Well-known member
MHB Site Helper
Hello !
I have to find the orthogonal trajectories of the curves :

$$x^{2}-y^{2}=c , x^{2}+y^{2}+2cy=1$$ ..

How can I do this??

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

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

Klaas van Aarsen

MHB Seeker
Staff member
Hello !
I have to find the orthogonal trajectories of the curves :

$$x^{2}-y^{2}=c , x^{2}+y^{2}+2cy=1$$ ..

How can I do this??

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

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

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

What do you want to continue with?

evinda

Well-known member
MHB Site Helper
Hi evinda!

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

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

Klaas van Aarsen

MHB Seeker
Staff member
Since both of the curves are given at the same subquestion,don't I have to find something common between those two solutions?? Or,what do I have to do??
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
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
Hold on!

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).

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.

evinda

Well-known member
MHB Site Helper
Hold on!

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 Thank you very much!!!