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

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,

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

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.