How would one go about proving that the major and minor axes of an ellipse are perpendicular bisectors of each other? I've read several proofs of expressions for the ellipse in cartesian and polar coordinates given the definition of the ellipse as the collection of points where the sum of the distance from two points is constant, but they always just assume that it is true. Visually it looks true, but is there an argument that follows simply from the definition?
[Math] Orthogonality of major and minor axes of an ellipse
conic sectionsgeometry
Related Solutions
More explicitly, we have the decomposition
$$\begin{pmatrix}\cos\,t\\\cos\,t+\sin\,t\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}\cos(t+\eta)\\\sqrt{2-\phi}\sin(t+\eta)\end{pmatrix}$$
where $\tan\,\lambda=\phi$, $\tan\,\eta=1-\phi$, and $\phi=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio. You can check that both your original parametric equations and the new decomposition both satisfy the Cartesian equation $2x^2-2xy+y^2=1$. What the decomposition says is that your curve is an ellipse with axes $\sqrt{1+\phi}$ and $\sqrt{2-\phi}$, with the major axis inclined at an angle $\lambda$.
If we take the linear algebraic viewpoint, as suggested by Robert in the comments, what the decomposition given above amounts to is the singular value decomposition (SVD) of the shearing matrix; i.e.,
$$\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}\cos\,\lambda&-\sin\,\lambda\\\sin\,\lambda&\cos\,\lambda\end{pmatrix}\cdot\begin{pmatrix}\sqrt{1+\phi}&\\&\sqrt{2-\phi}\end{pmatrix}\cdot\begin{pmatrix}\cos\,\eta&\sin\,\eta\\-\sin\,\eta&\cos\,\eta\end{pmatrix}^\top$$
The SVD is in fact an excellent way to look at how a matrix transformation geometrically affects points: the two orthogonal matrices on the left and right can be thought of as rotation matrices, reflection matrices, or products thereof, and the diagonal matrix containing the singular values amounts to nothing more than a scaling about the axes of your coordinate system.
If you are looking for the length of the minor and major axis you can calculate $ r_{min} $ and $ r_{max} $ (see formulae below).
If you are trying to determine the bounding box, you can calculate the left-most, right-most, top-most and bottom-most points.
As far as the angle of rotation is concerned, I use the algorithm and formulae below.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form: $ a x^2 + b x y + c y^2 + d x + e y + f = 0 $.
The equation represents an ellipse if $ b^2 - 4 a c < 0 $ , or similarly, $ 4 a c - b^2 > 0 $
The coefficient normalizing factor is given by:
$ q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}} $
The distance between center and focal point (either of the two) is given by:
$ s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }} $
The semi-major axis length is given by:
$ r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }} $
The semi-minor axis length is given by:
$ r_\min = \sqrt {{r_\max}^2 - s^2} $
The center of the ellipse is given by:
$ x_\Delta = { b e - 2 c d \over 4 a c - b^2} $
$ y_\Delta = { b d - 2 a e \over 4 a c - b^2} $
The top-most point on the ellipse is given by:
$ y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_T = {{-b y_T - d} \over {2 a}} $
The bottom-most point on the ellipse is given by:
$ y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_B = {{-b y_B - d} \over {2 a}} $
The left-most point on the ellipse is given by:
$ x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_L = {{-b x_L - e} \over {2 c}} $
The right-most point on the ellipse is given by:
$ x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_R = {{-b x_R - e} \over {2 c}} $
The angle between x-axis and major axis is given by:
if $ (q a - q c = 0) $ and $ (q b = 0) $ then $ \theta = 0 $
if $ (q a - q c = 0) $ and $ (q b > 0) $ then $ \theta = {1 \over 4} \pi $
if $ (q a - q c = 0) $ and $ (q b < 0) $ then $ \theta = {3 \over 4} \pi $
if $ (q a - q c > 0) $ and $ (q b >= 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} $
if $ (q a - q c > 0) $ and $ (q b < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi} $
if $ (q a - q c < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi} $
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Best Answer
As a very simple argument you can consider that an ellipse could be obtained by stretching a circle along a diameter, thus major and minor axes of an ellipse are perpendicular.
The reason for this is very clear considering the canonical form of an ellipse: $$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$$ which can be transformed in a circle stretching one axis.
EG $$z=\frac{b}{a} y \to x^2+ z^2=a^2$$
This is also the reason for the formula to calculate the area of the ellipse $A=\pi ab$.