Find the center of an ellipse such that it is tangent to a parabola at a point on it

Question

Question:

Given the parabola

$ (r - V)^T R D_p R^T (r - V) + b_0^T R^T (r - V) = 0 $

where $r = [x, y]^T $, $D_p = \begin{bmatrix} a && 0 \\ 0 && 0 \end{bmatrix} $, $R$ is a $2 \times 2 $ rotation matrix, $b_0 = [0, -1]^T $, and $V$ is the vertex of the parabola. And let $r_1$ be a point on the parabola. In, let an ellipse be given by

$ (r - C)^T Q (r - C) = 1 $

Determine $C$ such that the ellipse is tangent to the parabola at $r_1$, from the outside.

A similar problem involving two ellipses is posted here.

---

My attempt:

At $r_1$ the normal vector to the parabola and pointing outward is

$ n = 2 R D_p R^T (r_1 - V) + R b_0 $

Note that $n$ is a known vector, because all the quantities involved are known.

On the other hand, the negative of $ n$ is the direction along which the normal to the ellipse is pointing, so

$ - K n = Q (r_1 - C) $

So that

$ (r_1 - C) = - K Q^{-1} n $

Substituting this into the ellipse equation, gives

$ K = \dfrac{1}{\sqrt{ n^T Q^{-1} n } } $

Therefore,

$ C = r_1 + \dfrac{ Q^{-1} n } { \sqrt{ n^T Q^{-1} n } } $

Answer 1

Let's take $a = 0.25, V = (5, 7), R = R(\theta)$, with $\theta = - \dfrac{\pi}{6} $. This specifies the parabola. These parameters correspond to a parametric form

$ p(t) = \begin{bmatrix} 5 \\ 7 \end{bmatrix} + \begin{bmatrix} \cos 30^\circ && \sin 30^\circ \\ -\sin 30^\circ && \cos 30^\circ \end{bmatrix} \begin{bmatrix} t \\ 0.25 \ t^2 \end{bmatrix} $

Next, take $ t = 2 $, then

$ r_1 = p(1) = (7.232, 6.866) $

The normal vector to the parabola curve and pointing outward is

$ n = 2 R D_p R^T (r_1 - V) + R b_0 $

This evaluates to

$ n = (0.366025, -1.36603)$

Now using the formulas above, we obtain

$ C = (7.749, 4.934) $

The figure below shows the parabola, and the tangent ellipse.