# Angle between a Point on a Ellipse and its Center

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?

#### Attachments

Nidum
Gold Member
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

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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?

Nidum
Gold Member
A vertical line dropped from point B cuts the AO line ?

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

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

Nidum
Gold Member
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.

Nidum
Gold Member
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 ?

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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.

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Nidum