Tucking an ellipse between two rays

Let $\theta_1 = - 30^{\circ} = -\dfrac{\pi}{6}$ and $\theta_2 = 45^{\circ} = \dfrac{\pi}{4}$ . Define the unit vectors along the rays and perpendicular to the rays as

$u_1 = \begin{bmatrix} \cos \theta_1 \\ \sin \theta_1 \end{bmatrix} , u_2 = \begin{bmatrix} \sin \theta_1 \\ -\cos \theta_1 \end{bmatrix}, u_3 = \begin{bmatrix} \cos \theta_2 \\ \sin \theta_2 \end{bmatrix}, u_4 = \begin{bmatrix} -\sin \theta_2 \\ \cos \theta_2 \end{bmatrix}$

Note that the perpendicular vectors $u_2$ and $u_4$ are pointing outward from the region that is bounded in between the two rays.

The equation of our ellipse is $( r - C )^T R D R^T (r - C) = 1$ , where the diagonal matrix $D = \text{diag} \{ \dfrac{1}{20^2} , \dfrac{1}{11^2} \}$ , and matrix $R$ is a rotation matrix by an angle of $\phi = 75^{\circ} = \dfrac{5 \pi}{12}$ and is given by

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

And $r$ is the coordinate vector of a point on the ellipse, while $C$ is the coordinate vector of the center of the ellipse.

The gradient vector which points normal to the ellipse in an outward direction from the ellipse is given by $\nabla(r) = 2 R D R^T (r - c)$ . At the tangent points with the two rays we have

$R D R^T (r_1 - C) = \alpha u_2$ and $R D R^T (r_2 - C) = \beta u_4$ , where $\alpha \gt 0 , \beta \gt 0$ .

Solving for $(r_1 - C)$ , we obtain, $r_1 - C = \alpha R D^{-1} R^T u_2$ and similarly $r_2 - C = \beta R D^{-1} R^T u_4$

Substituting these into the equation of the ellipse, we deduce that

$\alpha = \dfrac{1}{ \sqrt{ u_2^T R D^{-1} R^T u_2 } }$

$\beta = \dfrac{1}{ \sqrt{ u_4^T R D^{-1} R^T u_4 }}$

Now that $\alpha$ and $\beta$ are known, we can proceed to find the center $C$ and the tangent points $r_1$ and $r_2$ .

Here we note that $r_1$ is along the vector $u_1$ , therefore, by premultiplying by the perpendicular vector it is eliminated.

$u_2^T (r_1 - C ) = 0 - u_2^T C = \alpha u_2^T R D^{-1} R u_2 = \sqrt{ u_2^T R D^{-1} R^T u_2 }$

Where the right hand side of the above equation is known. Similarly, we can write a second linear equation in the unknown vector $C$ , for the the second ray,

$u_4^T (r_2 - C ) = 0 - u_4^T C = \beta u_4^T R D^{-1} R u_4 = \sqrt{ u_4^T R D^{-1} R^T u_4 }$

So that now we have the following linear system

$A C = b$

where the matrix A is given by

$A = \begin{bmatrix} - u_2^T \\ - u_4^T \end{bmatrix}$

and the vector $b$ is

$b = \begin{bmatrix} \sqrt{ u_2^T R D^{-1} R^T u_2 } \\ \sqrt{ u_4^T R D^{-1} R^T u_4 } \end{bmatrix}$

And this can solved readily for the unknown coordinate vector of the center $C$ .

Having found the center, the tangent points $r_1$ and $r_2$ can be found immediately from the expressions above.

Namely,

$r_1 = C + \alpha R D^{-1} R^T u_2$

and

$r_2 = C + \beta R D^{-1} R^T u_4$

This concludes our calculations. The required sum comes to $88.473$ , and this makes the answer $\boxed{ 88473}$ .

Another solution is possible by considering a reference frame having its origin $O'$ at the center of the ellipse and with its $x'$ axis along the major axis of the ellipse and its $y'$ axis along the minor axis. Since the ellipse is rotated by $75^{\circ}$ counterclockwise with respect to the global reference frame, then the two tangent rays are rotated by $- 75^{\circ}$ with respect to its local reference frame $O' x' y'$ . Therefore, the direction vectors of the normals to these rays pointing outward of the ellipse are simply $R^T u_2$ and $R^T u_4$ , where the matrix $R$ is defined above, and premultiplying by its transpose has the effect of rotating $u_2$ and $u_4$ by $-75^{\circ}$ ( i.e. $75^{\circ}$ clockwise ). Now the equation of the ellipse with respect to its local reference frame is $r'^T D r' = 1$ , so that its gradient (which is normal to it) is $\nabla(r') = 2 D r'$ , and this means that at the two tangent points, we have, respectively,

$r'_1 = \alpha D^{-1} R^T u_2$ and $r'_2 = \beta D^{-1} R^T u_4$ , $\alpha \gt 0 , \beta \gt 0$

So that, upon substituting this into the equation of the ellipse, yields the same expressions for $\alpha$ and $\beta$ as above. Thus , we now have the tangent point $r'_1$ and $r'_2$ . What is left is to find the intersection point $P_0$ between the two tangents at $r_1$ and $r_2$ . And for that we can write the linear system in the parameters $t$ and $s$ as follows

$P_0 = r'_1 + t R^T u_1 = r'_2 + s R^T u_3$

and this can be solved for $t$ and $s$ , and thus we've obtained $P_0$ .

At the final step, and since the coordinates required are in the global frame, we have to relate the two frames $O' x' y'$ and the global frame $O xy$ . Note the point $P_0$ is the origin of the global system, and that the global frame is rotated by $R^T$ with respect to the local frame. The relation between the two frames is

$r' = P_0 + R^T r$

so that,

$r = R ( r' - P_0 )$

And this concludes the second method.

One can see that the results will be identical, if one considers the solution for $P_0$ in the second method. We have,

$P_0 = r'_1 + t R^T u_1 = r'_2 + s R^T u_3$

Premultiplying by $R$ through, we obtain,

$R P_0 = R r'_1 + t u_1 = R r'_2 + s u_3$

Premultiplying by $u_2^T$ ,

$u_2^T R P_0 = u_2^T R r'_1 = \alpha u_2^T R D^{-1} R^T u_2 = \sqrt{ u_2^T R D^{-1} R^T u_2 }$

similarly,

$u_4^T R P_0 = \sqrt{ u_4^T R D^{-1} R^T u_4 }$

Writing the above two equations as a linear system, and comparing with the linear system obtained in the first method, one deduces that $C = - R P_0$ . In addition, $R (r'_1 - P_0 ) = R r'_1 + C = r_1$ and $R(r'_2 - P_0) = R r'_2 + C = r_2$ .