Shifting an Ellipse to Tangency

Geometry Level 3

The above figure shows two ellipses. The first one (in red) is centered at v 0 = ( 3 , 4 ) \mathbf{v}_0 = (3, 4) and its semi-axes are congruent with the vectors v 1 = ( 4 , 1 ) \mathbf{v}_1 = (4, 1) and v 2 = ( − 1.5 , 6 ) \mathbf{v}_2 = (-1.5, 6) . The second one (in blue) is centered at w 0 = ( 6 , 5 ) \mathbf{w}_0 = (6, 5) and its semi-axes are congruent with the vectors w 1 = ( 3 , − 2 ) \mathbf{w}_1 = (3, -2) and w 2 = ( 3 , 4.5 ) \mathbf{w}_2 = (3, 4.5) . We want to shift the blue ellipse along the direction vector d = ( 1 , 2 ) \mathbf{d} = (1, 2) such that it becomes tangent to the red ellipse. This is accomplished by moving its center to a new location w 0 ∗ = w 0 + t ∗ d ^ {\mathbf{w}_0}^* = \mathbf{w}_0 + t^* \hat{\mathbf{d} } , where t ∗ > 0 t^* \gt 0 and d ^ \hat{\mathbf{d} } is the unit vector along d \mathbf{d} ; that is, d ^ = d / ∣ d ∣ = ( 1 5 , 2 5 ) \hat{\mathbf{d} } =\mathbf{d} / |\mathbf{d}| = (\dfrac{1}{\sqrt{5}}, \dfrac{2}{\sqrt{5}}) . Find the value of t ∗ t^* , which is the distance travelled by the center of blue ellipse to the final location, rounded to 3 decimal places. Note that the semi-axes orientations are kept unchanged. The figure below shows the final relative position of both ellipses.


The answer is 7.523.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Hosam Hajjir
Jul 14, 2018

Let r \mathbf{r} be a point on the first ellipse, then

r = v 0 + V u \mathbf{r} = \mathbf{v}_0 + V \mathbf{u} , where V = [ v 1 , v 2 ] V = [ \mathbf{v}_1, \mathbf{v}_2 ] and u \mathbf{u} is a unit vector , i.e. u = [ cos ⁡ θ , sin ⁡ θ ] T \mathbf{u}= [ \cos \theta , \sin \theta ]^T .

From this, it follows that,

u = V − 1 ( r − v 0 ) \mathbf{u} = V^{-1} (\mathbf{r} - \mathbf{v}_0) .

Upon imposing the unit length condition of vector u \mathbf{u} , the familiar equation of an ellipse results, namely,

( r − v 0 ) T Q 1 ( r − v 0 ) = 1 (\mathbf{r} - \mathbf{v}_0)^T Q_1 (\mathbf{r} - \mathbf{v}_0) = 1

where Q 1 = V − T V − 1 Q_1 = V^{-T} V^{-1}

Similarly, the algebraic equation of the second ellipse is

( r − ( w 0 + t d ^ ) ) T Q 2 ( r − ( w 0 + t d ^ ) ) = 1 (\mathbf{r} - (\mathbf{w}_0 + t \hat{\mathbf{d}}) )^T Q_2 (\mathbf{r} - (\mathbf{w}_0 + t \hat{\mathbf{d}}) ) = 1

where, Q 2 = W − T W − 1 Q_2 = W^{-T} W^{-1} , and W = [ w 1 , w 2 ] W = [ \mathbf{w}_1, \mathbf{w}_2 ] .

At tangency, since the points on the first ellipse are external to the second ellipse, except for a single point which is the point of tangency, we have

( v 0 + V u − ( w 0 + t d ^ ) ) T Q 2 ( v 0 + V u − ( w 0 + t d ^ ) ) ≥ 1 (\mathbf{v}_0+V \mathbf{u} - (\mathbf{w}_0 + t \hat{\mathbf{d}}) )^T Q_2 (\mathbf{v}_0+V \mathbf{u} - (\mathbf{w}_0 + t \hat{\mathbf{d}}) ) \ge 1

The left-hand side of the equation is a function of θ \theta and t ∗ t^* , and at the point of tangency this function becomes equal to 1, and this

is its minimum, and hence, at this point, its derivative with respect to θ \theta is zero. Hence, we have two scalar equations in two unknowns: θ ∗ \theta^* and t ∗ t^* . These are,

( v 0 + V u ∗ − ( w 0 + t ∗ d ^ ) ) T Q 2 ( v 0 + V u ∗ − ( w 0 + t ∗ d ^ ) ) = 1 (\mathbf{v}_0+V \mathbf{u}^* - (\mathbf{w}_0 + t^* \hat{\mathbf{d}}) )^T Q_2 (\mathbf{v}_0+V \mathbf{u}^* - (\mathbf{w}_0 + t^* \hat{\mathbf{d}}) ) = 1

and

( v 0 + V u ∗ − ( w 0 + t ∗ d ^ ) ) T Q 2 V d u d θ ∗ = 0 (\mathbf{v}_0+V \mathbf{u}^* - (\mathbf{w}_0 + t^* \hat{\mathbf{d}}) )^T Q_2 V {\dfrac{d \mathbf{u}}{d \theta}}^* = 0

One convenient way to solve this system of equations is by the multi-variate Newton-Raphson method. The algorithm is very fast, and in the case of this problem, it converged to the solution in 5 iterations. The solution vector was ( θ ∗ , t ∗ ) = ( 0.52485 , 7.52345 ) ( \theta^*, t ^*) = (0.52485, 7.52345) . Therefore, the answer is 7.523 \boxed{7.523} .

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