Determine ellipse from two points and direction vectors at those points

Question

I'm attempting to draw an ellipse based on two points. For each of these points I have a vector showing the direction the curve of the ellipse should be at this point (I suppose another way of looking at it is that I have 2 tangent lines to the to the ellipse with the intersection point for each).

I know normally 2 points would not be enough to determine the ellipse, but I thought the vectors might make it possible. Ideally I'm trying to calculate the center point and the major and minor axis (I guess either the actual points or the vector from the center).

I'm not quite sure how to proceed with this, or whether it's actually possible, but any help would be greatly appreciated, thanks.

Edit: Added a simple example of what I'm talking about. For the record, the calculation will be in 3D space.

Okay, so the I have the two tangent lines illustrated here, in the form of a point and a (normalized) vector, the point being the intersection point with the ellipse (it's part of the line and the ellipse). By vectors from the center point for the axes I just meant it as an alternative way of finding the minor and major axis points, stupid thing to put in).

Edit 2: Let's assume they're not parallel.

Answer 1

An ellipse is a conic, and a conic is uniquely defined by five points, no four of which lie on a common line. An ellipse is not a very special conic, so while some sets of five points may yield hyperbolas instead, you still have five real degrees of freedom, associated with these points.

Now you can consider the situation where two of the defining points of intersection are the touching points of your tangents, while another pair of two moves along these tangents towards the touching points. In the limit, when the moving points arrive at the touching points, this will result in the tangents you require.

But this still only covers four points, so there is still a one-parameter family of conics satisfying your requirements, corresponding to the one real degrees of freedom represented by the choice of a fifth point in the plane, modulo the movement of said point in a direction tangential to the conic it defines.

Answer 2

As mentioned in the comment, we need five points to determine a conic. Since a point together with a direction is equivalent to two points, two points plus two directions is not enough to uniquely determine a conic. we need at least one more point or some other condition.

I can imagine in some application, one will prefer to find an ellipse as "round" as possible. In following paragraphs, I'll outline how to find the one with minimal eccentricity.

Let $P_1$ and $P_2$ be the two points we want the ellipse to pass through. Let $\ell_1$ be the line we want the ellipse to have as a tangent line at $P_1$. Define $\ell_2$ in a similar manner. We will assume $\ell_1 \not\parallel \ell_2$. As a result, $\ell_1$ and $\ell_2$ intersect at some unique point $O$. We will choose the coordinate system such that $O$ is the origin.

Let $\vec{p}_1$ and $\vec{p}_2$ be the vectors $\overrightarrow{OP}_1$ and $\overrightarrow{OP}_2$. Since $\ell_1 \not\parallel \ell_2$, $\vec{p}_1$ and $\vec{p}_2$ form a basis of $\mathbb{R}^2$. For any point $X$ in the plane, we can represent the corresponding vector $\vec{x} = (x,y) \stackrel{def}{=} \overrightarrow{OX}$ as

$$\vec{x} = u \vec{p}_1 + v\vec{p}_2$$

As mentioned in the comment, there are infinitely many ellipses passing through $P_1$ and $P_2$ having $\ell_1$ and $\ell_2$ as tangent lines. In terms of $u, v$, this family of ellipses all have the form

$$(u-1)^2 + (v-1)^2 + 2\alpha u v = 1\quad\text{ with }\quad |\alpha| < 1\tag{*1}$$

To convert this to an equation in $\vec{x}$ and hence $(x,y)$, let $\vec{q}_1, \vec{q}_2$ be the dual basis for the basis $\vec{p}_1$, $\vec{p}_2$ of $\mathbb{R}^2$. More precisely, $\vec{q}_1$ and $\vec{q}_2$ are the two vectors defined by the relation:

$$\vec{p}_1 \cdot \vec{q}_1 = \vec{p}_2 \cdot \vec{q}_2 = 1 \quad\text{ and }\quad \vec{p}_1 \cdot \vec{q}_2 = \vec{p}_2 \cdot \vec{q}_1 = 0 $$ It is easy to see $$\vec{x} = u \vec{p}_1 + v \vec{q}_1 \quad\iff\quad \begin{cases}u = \vec{q}_1\cdot \vec{x},\\ v = \vec{q}_2\cdot\vec{x}\end{cases}$$ and $(*1)$ can be rewritten as

$$(\vec{q}_1\cdot \vec{x} - 1)^2 + (\vec{q}_2\cdot \vec{x} - 1)^2 + 2\alpha(\vec{q}_1\cdot \vec{x})(\vec{q}_2\cdot \vec{x}) = 1\tag{*2}$$

The LHS is a quadratic polynomial in $x$ and $y$. It is well known the center of corresponding ellipse, $\vec{x}_c$, coincides with the critical point of this quadratic polynomial. Taking gradient of LHS with respect to $\vec{x}$, we get

$$ \vec{q}_1 ( \vec{q}_1 \cdot \vec{x}_c + \alpha \vec{q}_2\cdot \vec{x}_c - 1 ) + \vec{q}_2 ( \vec{q}_2 \cdot \vec{x}_c + \alpha \vec{q}_1\cdot \vec{x}_c - 1 ) = \vec{0}$$ As a result, $$ \vec{q}_1 \cdot \vec{x}_c = \vec{q}_2 \cdot \vec{x}_c = \frac{1}{1+\alpha} \quad\implies\quad \vec{x}_c = \frac{1}{1+\alpha}(\vec{p}_1 + \vec{p}_2) $$

Let $M$ be the mid-point of $P_1P_2$. Independent of $\alpha$, the center $\vec{x}_c$ always lies on the line joining $O$ and $M$. Let's see how the ellipses look like in various limits.

• If $\alpha \sim -1$, the center $\vec{x}_c$ will be far away. The corresponding ellipse will be large and long with semi-major axis roughly in the direction of $\vec{p}_1 + \vec{p}_2$. The eccentricity of the ellipse $e$ will be very high. If we increase $\alpha$ away from $-1$, the eccentricity start to drop.

• If $\alpha \sim 1$, the center $\vec{x}_c$ will be close to the mid-point $M$ of $P_1P_2$. The corresponding ellipse will be short and looks like a line segment between $P_1$ and $P_2$. Once gain, the eccentricity will be very high. As we decrease $\alpha$ away from $1$, the eccentricity drop again.

• If $\alpha = 0$, $\vec{x}_c = \vec{p}_1+\vec{p}_2$. The center now coincides to the point $N$ where $OP_1NP_2$ forms a parallelogram. The eccentricity of this ellipse is usually moderate.

Based on this observation, the decision to use a particular $\alpha$ is equivalent to picking a point on a ray starting at $M$ pointing in the direction of $N$ and use that point as the center $\vec{x}_c$ of the ellipse.

Among all these ellipses, there is one who eccentricity $e$ is minimized. To determine the corresponding $\alpha$, let us first rewrite $(*2)$ to a from symmetric with respect to $\vec{x}_c$.

$$(\vec{q}_1\cdot ( \vec{x} - \vec{x}_c) )^2 + (\vec{q}_2\cdot ( \vec{x} - \vec{x}_c)^2 + 2\alpha(\vec{q}_1\cdot (\vec{x} - \vec{x}_c) )(\vec{q}_2\cdot ( \vec{x} - \vec{x}_c) ) = \frac{1-\alpha}{1+\alpha}$$

We will make a further assumption that

$$\begin{cases}\vec{p}_1 = (1,0),\\ \vec{p}_2 = (r,s)\end{cases} \quad\iff\quad \begin{cases}\vec{q}_1 = (1, -\frac{r}{s}),\\ \vec{q}_2 = (1, \frac{1}{s})\end{cases} $$ and $s > 0$. If we let $\vec{x} - \vec{x}_c = (\tilde{x}, \tilde{y})$, we get

$$\left(\tilde{x} - \frac{r}{s} \tilde{y}\right)^2 + \left(\frac{\tilde{y}}{s}\right)^2 + 2\alpha \left(\tilde{x} - \frac{r}{s} \tilde{y}\right)\left(\frac{\tilde{y}}{s}\right) = \frac{1-\alpha}{1+\alpha} $$ We can recast this into a matrix form

$$\begin{bmatrix}\tilde{x}\\ \tilde{y}\end{bmatrix}^T \begin{bmatrix} 1 & \frac{\alpha-r}{s}\\ \frac{\alpha-r}{s} & \frac{r^2-2\alpha r+1}{s^2}\end{bmatrix} \begin{bmatrix}\tilde{x} \\ \tilde{y}\end{bmatrix} = \frac{1-\alpha}{1+\alpha} $$ Let $\Delta$ be the $2\times 2$ matrix in above expression. Its characteristic polynomial has the form $$\det(\lambda I_2 - \Delta ) = \lambda^2 - (1 + \frac{r^2-2\alpha r+1}{s^2}) \lambda + \frac{1-\alpha^2}{s^2} $$ Let $\lambda_1 > \lambda_2$ be the two eigenvalues of $\Delta$. Let $a$ and $b$ be the semi-major axis and semi-minor axis for the ellipse. It is easy to see

$$ \lambda_1 b^2 = \lambda_2 a^2 = \frac{1-\alpha}{1+\alpha} \quad\implies\quad a^2 : b^2 = \lambda_1 : \lambda_2 \quad\iff\quad e^2 = \frac{a^2 - b^2}{a^2} = 1 - \frac{\lambda_2}{\lambda_1} $$ To minimize $e$, we need to maximize the ratio $\displaystyle\;\frac{\lambda_2}{\lambda_1}\;$. This is equivalent to maximizing the expression

$$\frac{\lambda_1\lambda_2}{(\lambda_1+\lambda_2)^2} = \frac{\det(\Delta)}{\text{Tr}(\Delta)^2} = \frac{\frac{1-\alpha^2}{s^2}}{\left( 1 + \frac{r^2-2\alpha r+1}{s^2}\right)^2} $$ Taking logarithm and differentiate, this condition for maximizing above expression becomes $$\frac{-2\alpha}{1-\alpha^2} = 2\left(\frac{-2r}{r^2 - 2\alpha r + 1 + s^2}\right) \quad\implies\quad \alpha_{min} = \frac{2r}{r^2+s^2+1} $$ We can rewrite the last formula only using vector operations

$$ \bbox[8pt,border:1px solid blue;]{ \alpha_{min} = \frac{2\vec{p}_1\cdot\vec{p}_2}{|\vec{p}_1|^2 + |\vec{p}_2|^2} } $$ It is now in a form independent of choice of coordinates and remain valid even when the points $P_1$ and $P_2$ live in $\mathbb{R}^3$.

As an example, consider the special case $\ell_1 \perp \ell_2$. We have

$$\vec{p}_1 \cdot \vec{p}_2 = 0 \quad\implies\quad \alpha_{min} = 0 \quad\implies\quad \vec{x}_c = \vec{p}_1 + \vec{p}_2.$$ The center of the minimal ellipse is the point $N$ we encountered before. Up to ordering, the semi-major and semi-minor axis of the ellipse is parallel to $\ell_1$ and $\ell_2$.