# Angle between a Point on a Ellipse and its Center

• ecastro
In summary, the problem statement is finding the angle between a point on the ellipse and the ellipse's center, and given/known data is that the distance between point O and point A is ##d## and the angle a is outside the ellipse between the lines AO and AB. The ellipse has a major-axis of ##r_1## and a minor axis of ##r_2##. The angle needed is ##x##, the angle between the lines AO and BO, inside the ellipse. However, the line AB is not tangent to the ellipse. The equation which defines an ellipse is ##\frac{x^2}{a^2}+
ecastro
1. The problem statement, aall variables, and given/known data
I need to find the angle between a point on the ellipse and the ellipse's center shown in the figure below:

The known variables are 'd', the distance between point O (the center of the ellipse) and point A, and 'a', the angle, outside the ellipse, between the lines AO and AB. The ellipse has a major-axis of ##r_1## and a minor axis of ##r_2##. The needed angle is 'x', the angle between the lines AO and BO, inside the ellipse. Note, however, that the line AB is not tangent to the ellipse.

Edit: By the way, 'd' is equal to the sum of ##H## and ##r_1##.

## Homework Equations

The relevant equations might be sines and cosines or even the sine and cosine laws. I am not sure since the line AB is not tangent to the ellipse.

## The Attempt at a Solution

I tried calculating the line perpendicular to AO to line AB (the opposite side of 'a' shown in the figure), but I do not know how to proceed then. The only solution I can think of is putting everything in a coordinate system and find the intersection of the line AB and the ellipse. Is there a more elegant solution than this one?

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What is the equation which defines an ellipse ?

Can you see a way to relate the x y coordinates of point B to the angle X ?

Take point O as the origin of coordinates

nb : Would be a good idea to call the angle something other than X to avoid confusion with the coordinate x

Last edited:
Nidum said:
Hint 1 : What is the equation which defines an ellipse ?

Hint 2 : Can you see a way to relate the x y coordinates of point B to the angle x ?

Take point O as the origin of coordinates

I could indeed use the equation of an ellipse, ##\displaystyle{\frac{x^2}{r_1^2} + \frac{y^2}{r_2^2} = 1}##, and then using the line ##y = 0##, which is a horizontal line, and have another line ##y_1 = mx_1 + b## which intersects the line ##y = 0## and solve the equation of the line having an angle with ##y = 0## equal to ##a##, and then find the point where this line intersects the ellipse, and I could use the Pythagorean theorem and trigonmetry to solve the angle.

But I want to know, is there a more elegant solution than this one?

A vertical line dropped from point B cuts the AO line ?

Nidum said:
A vertical line dropped from point B cuts the AO line ?

Yes, and it will be perpendicular to the AO line.

If we call the point of intersection point E then :

(1) Line BE is common to two triangles and is of known length for any angle X ?

(2) For any angle X you have the lengths EO and AE ?

The length of BE is not known, and I don't know how to solve it because I do not know the length of line AE.

Length BE is just the perpendicular distance from the point B to line AO . For any angle X you know the y coordinate of point B ?

Last edited:
The coordinate of point B is also not known.

As an update, I tried calculating the point of intersection between the line ##\overline{AB}## and the ellipse, and it turns out that it is more complicated than I thought. Here is my attempt:

Letting the semi-major axis of the ellipse be ##a## and the semi-minor axis be ##b##, the equation of the ellipse is
\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,
\end{equation*}

while the equation of the line intersecting the ellipse is
\begin{equation*}
y - y_0 = m\left(x - x_0\right) + y_\text{intercept}.
\end{equation*}

To find the point of intersection between the line and the ellipse, I will insert the equation of the line in the equation of the ellipse. So,
\begin{eqnarray*}
y &=& m\left(x - x_0\right) + y_\text{intercept} + y_0 \\
\Rightarrow \frac{x^2}{a^2} + \frac{\left[m\left(x - x_0\right) + y_\text{intercept} + y_0\right]^2}{b^2} &=& 1 \\
\Rightarrow \left[m\left(x - x_0\right) + y_\text{intercept} + y_0\right]^2 &=& m^2 \left(x - x_0\right)^2 \\
&+& 2 m y_\text{intercept} \left(x - x_0\right) + 2 m y_0 \left(x - x_0\right) \\
&+& y_\text{intercept}^2 + 2 y_\text{intercept} y_0 + y_0^2 \\
&=& m^2\left(x^2 - 2xx_0 + x_0^2\right) + 2mxy_\text{intercept} - 2mx_0y_\text{intercept} \\
&+& 2mxy_0 - 2mx_0y_0 + y_\text{intercept}^2 + 2y_\text{intercept}y_0 + y_0^2 \\
&=& m^2 x^2 - 2m^2xx_0 + m^2x_0^2 + 2mxy_\text{intercept} - 2mx_0y_\text{intercept} \\
&+& 2mxy_0 - 2mx_0y_0 + y_\text{intercept}^2 + 2y_\text{intercept}y_0 + y_0^2.
\end{eqnarray*}

Using the previous equation, the equation of the ellipse becomes
\begin{eqnarray*}
\frac{x^2}{a^2} + \frac{m^2 x^2 - 2m^2xx_0 + m^2x_0^2 + \cdots}{b^2} &=& 1.
\end{eqnarray*}

And then joining terms with ##x^2## and ##x## to form a polynomial function of order 2;
\begin{eqnarray*}
\left(\frac{1}{a^2} + \frac{m^2}{b^2}\right)x^2 + \left(\frac{-2m^2x_0 + 2my_\text{intercept} + 2my_0}{b^2} \right)x + \left(\frac{m^2x_0^2 - 2mx_0y_\text{intercept} - 2mx_0y_0 + y_\text{intercept}^2 + 2y_\text{intercept}y_0 + y_0^2}{b^2} - 1\right) = 0.
\end{eqnarray*}

This is similar to the form of the polynomial ##ax^2 + bx + c = 0##, which can be solved using the quadratic formula. Letting
\begin{eqnarray*}
a &=& \frac{1}{a^2} + \frac{m^2}{b^2}, \\
&=& \frac{b^2 + a^2 m^2}{a^2 b^2}, \\
b &=& \frac{-2m^2x_0 + 2my_\text{intercept} + 2my_0}{b^2}, \\
c &=& \frac{m^2x_0^2 - 2mx_0y_\text{intercept} - 2mx_0y_0 + y_\text{intercept}^2 + 2y_\text{intercept}y_0 + y_0^2}{b^2} - 1, \\
&=& \frac{m^2x_0^2 - 2mx_0y_\text{intercept} - 2mx_0y_0 + y_\text{intercept}^2 + 2y_\text{intercept}y_0 + y_0^2 - b^2}{b^2}.
\end{eqnarray*}

I plan to divide the solution to the quadratic equation into several parts for an easier reading of my attempt. The division of the quadratic equation will be ##-b##, ##b^2##, ##4ac##, and ##2a##.
\begin{eqnarray*}
-b &=& \frac{2mx_0 - 2my_\text{intercept} - 2my_0}{b^2}, \\
b^2 &=& \frac{\left(2m^2x_0 - 2my_\text{intercept} - 2my_0\right)^2}{b^4}, \\
&=& \frac{4 m^4 x_0^2 - 8 m^3 x_0 y_\text{intercept} - 8 m^3 x_0 y_0 + 4 m^2 y_\text{intercept}^2 + 8 m^2 y_\text{intercept} y_0 + 4 m^2 y_0^2}{b^4}, \\
4ac &=& 4 \left(\frac{b^2 + a^2 m^2}{a^2 b^2}\right) \left(\frac{m^2x_0^2 - 2mx_0y_\text{intercept} - 2mx_0y_0 + y_\text{intercept}^2 + 2y_\text{intercept}y_0 + y_0^2}{b^2} - 1\right), \\
&=& \frac{4 b^2 m^2 x_0^2 - 8 b^2 m x_0 y_\text{intercept} - 8 b^2 m x_0 y_0 + 4 b^2 y_\text{intercept}^2 + 8 b^2 y_\text{intercept} y_0 + 4 b^2 y_0^2 - 4 b^4 + 4 a^2 m^4 x_0^2 - 8 a^2 m^3 x_0 y_\text{intercept} - 8 a^2 m^3 x_0 y_0 + 4 a^2 m^2 y_\text{intercept}^2 + 8 a^2 m^2 y_\text{intercept} y_0 + 4 a^2 m^2 y_0^2 - 4 a^2 b^2 m^2}{a^2 b^4}, \\
2a &=& \frac{2 b^2 + 2 a^2 m^2}{a^2 b^2}.
\end{eqnarray*}

The next part will be ##b^2 - 4ac##;
\begin{eqnarray*}
b^2 - 4ac &=& \frac{4 m^4 x_0^2 - 8 m^3 x_0 y_\text{intercept} - 8 m^3 x_0 y_0 + 4 m^2 y_\text{intercept}^2 + 8 m^2 y_\text{intercept} y_0 + 4 m^2 y_0^2}{b^4} - \frac{4 b^2 m^2 x_0^2 - 8 b^2 m x_0 y_\text{intercept} - 8 b^2 m x_0 y_0 + 4 b^2 y_\text{intercept}^2 + 8 b^2 y_\text{intercept} y_0 + 4 b^2 y_0^2 - 4 b^4 + 4 a^2 m^4 x_0^2 - 8 a^2 m^3 x_0 y_\text{intercept} - 8 a^2 m^3 x_0 y_0 + 4 a^2 m^2 y_\text{intercept}^2 + 8 a^2 m^2 y_\text{intercept} y_0 + 4 a^2 m^2 y_0^2 - 4 a^2 b^2 m^2}{a^2 b^4} \\
&=& \left(\frac{4 a^2 m^4 x_0^2 - 8 a^2 m^3 x_0 y_\text{intercept} - 8 a^2 m^3 x_0 y_0 + 4 a^2 m^2 y_\text{intercept}^2 + 8 a^2 m^2 y_\text{intercept} y_0 + 4 a^2 m^2 y_0^2 - 4 b^2 m^2 x_0^2 + 8 b^2 m x_0 y_\text{intercept} + 8 b^2 m x_0 y_0 - 4 b^2 y_\text{intercept}^2 - 8 b^2 y_\text{intercept} y_0 - 4 b^2 y_0^2 + 4 b^4 - 4 a^2 m^4 x_0^2 + 8 a^2 m^3 x_0 y_\text{intercept} + 8 a^2 m^3 x_0 y_0 - 4 a^2 m^2 y_\text{intercept}^2 - 8 a^2 m^2 y_\text{intercept} y_0 - 4 a^2 m^2 y_0^2 + 4 a^2 b^2 m^2}{a^2 b^4}\right) \\
&=& \left(\frac{-4 b^2 m^2 x_0^2 + 8 b^2 m x_0 y_\text{intercept} + 8 b^2 m x_0 y_0 - 4 b^2 y_\text{intercept}^2 - 8 b^2 y_\text{intercept} y_0 - 4 b^2 y_0^2 + 4 b^4 + 4 a^2 b^2 m^2}{a^2 b^4}\right) \\
&=& \frac{-4b^2\left(m^2 x_0^2 - 2mx_0y_0 + y_0^2 - 2mx_0y_\text{intercept} + y_\text{intercept}^2 + 2y_\text{intercept}y_0 - b^2 - a^2m^2\right)}{b^4}.
\end{eqnarray*}

This is where I get stuck... I do not know how to simplify the equation further.

Last edited:
This problem is just a simple exercise in trig . I gave you the clue to getting the solution but you ignored it .

Nidum said:
This problem is just a simple exercise in trig . I gave you the clue to getting the solution but you ignored it .
I apologize for my ignorance; I never thought that it is a bit complicated, at least for me. I have edited my latest post (I accidentally posted it while I was doing a preview of the post).

## 1. What is the angle between a point on an ellipse and its center?

The angle between a point on an ellipse and its center is the angle formed by the line connecting the point and the center, and the major axis of the ellipse. This angle is also known as the eccentric angle or true anomaly.

## 2. How is the angle between a point on an ellipse and its center calculated?

The angle can be calculated using trigonometric functions such as sine, cosine, and tangent. The specific formula used will depend on the coordinates of the point and the shape of the ellipse.

## 3. Can the angle between a point on an ellipse and its center be negative?

Yes, the angle between a point on an ellipse and its center can be negative. This happens when the point is located on the opposite side of the major axis from the center, resulting in an angle greater than 180 degrees.

## 4. How does the angle between a point on an ellipse and its center change as the point moves along the ellipse?

The angle between a point on an ellipse and its center changes continuously as the point moves along the ellipse. It will start at 0 degrees when the point is located at the intersection of the major and minor axes, and will increase or decrease depending on the direction of movement along the ellipse.

## 5. What is the significance of the angle between a point on an ellipse and its center?

The angle between a point on an ellipse and its center is an important parameter in understanding the position of the point on the ellipse. It is also used in various equations and calculations involving ellipses, such as determining the distance between two points on an ellipse or the velocity of an object moving along an elliptical orbit.

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