Similar Tangent Ellipses

The equation of the original ellipse is

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

which can be written compactly as

$r^T D r = 1$

where $r = [x, y]^T$ , and

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

Now, for the rotated and scaled ellipse, the equation becomes

$\dfrac{1}{s^2} r^T R D R^T r = 1$

where $s$ is the scale factor, and

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

with $\theta = 30^{\circ}$ .

Let vector $p$ be a tangency point between the two ellipses, then $p$ satisfies the following

equations,

$p^T D p = 1 \text{ ............ (1) }$

$\dfrac{1}{s^2} p^T R D R^T p = 1 \text{ ............ (2) }$

In addition, from tangency, the gradient vector of both ellipses are parallel, therefore,

$D p = \alpha (\dfrac{1}{s^2} R D R^T p) \text{ ............ (3) }$

Substituting (3) in (1) and comparing with (2), it follows that $\alpha = 1$ , and hence, from (3), we now have,

$\left( \dfrac{1}{s^2} R D R^T - D \right) p = 0 \text{ ............ (4)}$

Since $p$ is a non-zero vector, then we must have,

$\left| \dfrac{1}{s^2} R D R^T - D \right| = 0 \text{ ............ (5)}$

The expansion of this $2 \times 2$ determinant yields a quadratic equation in $\dfrac{1}{s^2}$ . Since it is assumed that $s > 0$ , equation (5) gives two possible values of the scale factor $s$ . These correspond to two possible ellipses, one smaller and one larger than the original ellipse.

Since in this problem, we're interested in the smaller ellipse, we'll take $s < 1$ .

Performing the computations, yields $s = 0.693$ .