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Find orthogonal trajectories to the family of elipses: $$\frac{x^2}{a^2}+y^2=1$$ where $a>0$

The idea is to remove the parameter $a$ from the equation via clever integration so it drops out.

-> rearrange to get $y'=f(x,y)$

-> we go into the exuation of orthogonal trajectories. $$y'=\frac {-1}{f(x,y)}$$

-> solve this new DE to get the family of equations that give you orthogonal trajectories to the primary elipse.

BUT

I calcualted 3 times with 3 seperate solutions and at that point I decided that I better just ask someone.

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    You basically just need to differentiate the given equation of ellipse with respect to $x$ and rearrange to get an expression for $y'=\frac{dy}{dx}$. This might help: https://en.wikipedia.org/wiki/Orthogonal_trajectory – 2017-02-13

1 Answers 1

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$$ x^2/a^2 + y^2 =1 \tag1 $$

Differentiate wrt x

$$ x/a^2 + y y^\prime = 0 \tag2 $$

From 1) and 2) eliminate $1/a^2 $, simplify

$$ \frac{1-y^2}{x} = \frac{-y y^\prime}{1} \tag3 $$

which is the DE of all ellipses with unit semi minor axis

To get at orthogonal trajectory DE, $ y^\prime \rightarrow -1/ y^\prime $

$$ \frac{y^\prime }{y} = \frac{x}{1-y^2} \tag4 $$

Variable separable DE

$$ (1/y - y)dy = x\, dx \tag 5 $$

Integrate and simplify

$$ 2 log y = x^2+y^2 + c_1 \tag 6 $$

Simplify again to get $x=f(y)$ a relation or an implicit relation with arbitrary constants,

$$ x= \sqrt{ 2 \, log ( c_1\, y) - y^2 }, \quad y/e^ {\sqrt{ x^2+y^2}} = c_2 \tag7 $$

An plot for approximate loci indication

Please tell if any step is not clear, shall explain it.

EDIT1:

g1=ContourPlot[ y/ Exp[Sqrt[x^2+y^2]], {x,-4,4},{y,-1.1,1.1},AspectRatio-> .2]
g2=ContourPlot[ (1-y^2)/ x^2 , {x,-4,4},{y,-1.1,1.1},AspectRatio-> .2]
Show[{g1,g2}, PlotRange->All]
g5=ParametricPlot[ (1-y^2)/ x^2  , {x,-4,4},{y,-1.1,1.1},AspectRatio-> .5]

enter image description here

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    This makes perfect sense. But honestly I doupt I would have figured it out anytime soon as my thoughts went in completely the wrong direction before. Also thanks for the \tag# checked your source how you make those in markdown. Do wonder where you managed to do the plot so quickly. Cheers – 2017-02-13
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    Plotting better than ContourPlot is possible.Mathematica code is pasted in EDIT1 – 2017-02-13