Determining the cartesian equation of an ellipse ( be it neither in standard position nor orientation) given center, one vertex and one semi -axis

Question

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Question : I obtained the final formula while thinking of center $C$ and vertex $V$ as points lying in the first quadrant, with $V$ located to the right of $C$ Is the formula general in spite of this? - I assumed that the " shifting terms" $h$ and $k$ operate , so to say, in the rotated system, so that, intead of substituting $X_C$ for $h$ and $Y_C$ for $k$ , I used the coordinates of center $C$ in the rotated system. Is this assumption correct? Finally, can you think of other forms of this equation?

The givens are :

• $\text {Center} = C = (X_C, Y_C) $

• $ \text {Vertex} = V = ( X_V,Y_V) = (X_C+p, X_C+q) , \text {with }p, q$ $\in \mathbb R$

• $ m {\in\space \mathbb R^+} = \text {length of one semi axis ( perpendicular to the semi-axis [CV] }$

From what is given can be immediately deduced :

• $ M = \text {length of the semi axis [CV]}= \sqrt {p^2 + q^2}$

• $R = \text {inclination of the straight line (CV) on which lies the semi axis [CV] }= \arctan {\frac qp}$

• the equations of $X'$ and $Y'$, the rotated coordinate system in which lies the parabola , namely

$\space \space \space \space X' : y = \tan(R)x $

$\space \space \space \space Y' : y = (-1 / \tan(R))x $

Since the ellipse lies in a rotated coordinate system, and since its center is not necessarily at the origin its equation will be of the form :

$$\frac { (X(x,y) -h )^2 } { M^2 } + \frac { (Y(x,y)-k )^2 } { m^2 } =1$$

with

• $X(x,y) = x \cos(R) + y \sin (R) $

• $Y(x,y) = y \cos(R) - y \sin (R) $

• $ h$ and $k$ : the coordinates of center $C$ in the rotated coordinate system.

Determining the values of $h$ and $k$

• $h = X_C \cos (R) + Y_C \sin (R) $

• $k = Y_C \sin (R) - X_C \cos(R)$

Hence the formula for the ellipse of center $C= (X_C , Y_C) $, vertex $V$ and a semi-axis of length $m$ is :

$$\frac { (X(x,y) - (X_C \cos (R) + Y_C \sin (R)) )^2 } { M^2 } + \frac {( Y(x,y)-(Y_C \sin (R) - X_C \cos(R)) ) ^2 } { m^2 } =1$$

with, as said above $ M= \text {length of the semi- axis [CV]}$

Answer 1

Too big for a comment, but using the composition of arctan and sine/cosine actually simplifies things a lot more than I expected, because $M$ reappears. The identities are:

$$\sin \arctan \frac q p = \frac{\frac q p}{\sqrt{1+\left(\frac q p\right)^2}} = \frac p {\sqrt{p^2+q^2}} = \frac p M \\ \cos \arctan \frac q p = \frac{1}{\sqrt{1+\left(\frac q p\right)^2}} = \frac q {\sqrt{p^2+q^2}} = \frac q M$$

This makes your final equation at the bottom:

$$\frac{1}{M^2} \left( \left( \frac{p}{M}x + \frac{q}{M}y \right) - \left( \frac{p}{M}X_C + \frac{q}{M}Y_C \right) \right)^2 + \frac{1}{m^2} \left( \left( \frac{q}{M}y - \frac{p}{M}x \right) - \left( \frac{q}{M}X_C - \frac{p}{M}Y_C \right) \right)^2 = 1 \\ \frac{1}{M^4} \left( px + qy - pX_C - qY_C \right)^2 + \frac{1}{M^2m^2} \left( py -qx - pY_C + qX_C \right)^2 = 1 \\ \frac{1}{M^2} \left( px + qy - pX_C - qY_C \right)^2 + \frac{1}{m^2} \left( py -qx - pY_C + qX_C \right)^2 = M^2$$

I fear that further "simplification" just ends up with huge constants just to put it into $Ax^2+By^2+Cxy+Dx+Ey+F$ form.

Answer 2

Rotate then Shift.

Starting from the algebraic equation of an ellipse in standard format

$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 $

If you define the position vector $\mathbf{p} = [x, y]^T $ , then the above equation can be written concisely as follows

$ \mathbf{p}^T \ D \ \mathbf{p} = 1\hspace{50pt}(*)$

where

$ D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix} $

Now in the end, you want to find the algebraic equation of the same ellipse but rotated by some angle $\theta$ (with respect to the original orientation) and shifted by a certain displacement vector $(Xc, Yc)$

So, first rotate the ellipse about the origin, then shift the resulting the ellipse by the vector $(Xc , Yc)$

The image of a point $\mathbf{p} = (x,y)$ on the original ellipse under a rotation by an angle $\theta$ is the point

$\mathbf{p'} = R \ \mathbf{p} $

where $R$ is the two-dimensional rotation matrix given by

$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $

And the image of $\mathbf{p'} $ after a shift by $C = [X_c, Y_c]^T $ is

$ \mathbf{p''} = \mathbf{p'} + \mathbf{C} = R \ \mathbf{p} + \mathbf{C} $

From this last equation, we deduce that $ \mathbf{p} = R^T (\mathbf{p''} - \mathbf{C} ) $

Substituting this into equation $(*)$, gives us

$ (\mathbf{p''} - \mathbf{C})^T \ R D R^T \ (\mathbf{p''} - \mathbf{C} ) = 1 \hspace{50pt} (**)$

And this is the algebraic equation of the rotated/shifted ellipse.

Explicitly, we have

$ R D R^T = \begin{bmatrix} \dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta && \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta \\ \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta && \dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta \end{bmatrix}$

And $\mathbf{p''} - \mathbf{C} = (x - X_c, y - Y_c) $

Therefore, the algebraic equation is

$\bigg(\dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta \bigg) (x - X_c)^2 + 2 \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) \sin \theta \cos \theta (x - X_c) (y - Y_c) + \bigg(\dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta\bigg) (y - Y_c)^2 = 1 $

Using the double angle trigonometric formulas, this last equation can be expressed as follows

$(A + B \cos(2 \theta)) (x - X_c)^2 + 2 B \sin(2 \theta) (x - X_c) (y - Y_c) + (A - B \cos(2 \theta)) (y - Y_c)^2 = 1 $

where

$ A = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} + \dfrac{1}{b^2} \bigg) $

$ B = \dfrac{1}{2} \bigg( \dfrac{1}{a^2} - \dfrac{1}{b^2} \bigg) $