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How to get all the points in an ellipse when I know a point of it and it's center?

I have the following situation:

enter image description here

I know the position of the red dot relative to the center, I have the vertical distance and the horizontal distance values. What I need is to calculate all the other possible points in the ellipse, as to know what other points the dot can be if it orbits the ellipse.

I have a coordinate system to make that

How is it possible to be done?

@EDIT

I'm also able to know the major axis and the minor axis

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    $\begingroup$ You can't. You can draw an infinite number of ellipses with the same centre that pass through a given point. $\endgroup$ – copper.hat Jul 5 '16 at 22:59
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    $\begingroup$ What do you need it for? Maybe you have some additional information you did not consider? $\endgroup$ – Yuriy S Jul 5 '16 at 23:02
  • $\begingroup$ Generally an ellipse can be defined by two foci and a point lying on it, two foci and the sum of the distances from the foci for a point on the boundary, or by the center and its major and minor radii (and the direction of one of the radii). $\endgroup$ – florence Jul 5 '16 at 23:07
  • $\begingroup$ I'm able to know the horizontal radii and the vertical radii $\endgroup$ – Kerooker Jul 5 '16 at 23:22
  • $\begingroup$ @YuriyS title edited. I do have a coordinate system, but I don't know the equation x= $\endgroup$ – Kerooker Jul 5 '16 at 23:35
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Let $a,b$ be the major and minor axes of the ellipse.

Let $x_0,y_0$ be the coordinates of the center on the ellipse.

Let $x_1,y_1$ be the coordinates of the one point on the ellipse.

First, we write the equation of the ellipse, those major axis is parallel to the $x$ axis of the coordinate system (which is at this point completely arbitrary).

$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$

Now we need the ellipse to go through the point $x_1,y_1$. This requires first that the distance between this point and the center is somewhere between $a$ and $b$:

$$b \leq \sqrt{(x_1-x_0)^2+(y_1-y_0)^2} \leq a$$

If it's indeed true, we need to find the four corresponding points on the ellipse on the same distance from the center.

$$d=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$$

One of them is a green point on the picture below:

enter image description here

We need to find the green point. Let's call it $x_2,y_2$. Then we have two equations in two variables:

$$\frac{(x_2-x_0)^2}{a^2}+\frac{(y_2-y_0)^2}{b^2}=1$$

$$(x_2-x_0)^2+(y_2-y_0)^2=(x_1-x_0)^2+(y_1-y_0)^2$$

After you solve this simple system of equations and find $x_2,y_2$, there is one last step left: you need to rotate your ellipse.

First, find the angle between lines going from the center of the ellipse to $x_1,y_1$ and to $x_2,y_2$. You can do it by using vector dot product:

$$\cos \theta=\frac{(x_1-x_0)(x_2-x_0)+(y_1-y_0)(y_2-y_0)}{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}\sqrt{(x_2-x_0)^2+(y_2-y_0)^2}}$$

The rotated coordinates can be then found from the following system of equations:

$$x=X \cos \theta-Y \sin \theta \\ y=X \sin \theta+ Y \cos \theta$$

Note, that I inverted the problem - as if the ellipse is already rotated counter-clockwise, and we want to rotate it back. Meaning, we just substitute these formulas in the original equation:

$$\frac{(X \cos \theta-Y \sin \theta-x_0)^2}{a^2}+\frac{(X \sin \theta+ Y \cos \theta-y_0)^2}{b^2}=1$$

And here is the equation for your ellipse. You still need to find $x_2,y_2$ and $\cos \theta$ - I leave it to you.

See http://en.wikipedia.org/wiki/Rotation_matrix for more about rotation.

Note, that there is four possible green points you can choose, so there is four different equations you might need to check, not two.

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