Question P.172: Let ABCD be a parallelogram in the plane. Describe and sketc......

Solution 1. To show why this description is correct, choose coordinates in the plane such that A = (0, 0), B = (1, 0), and D = (λ, µ). Then C = (λ+ 1, µ). We may assume that µ > 0. For the four points (0, 0), (1, 0), (λ, µ), (λ + 1, µ) to lie on a conic section ax² + bxy + cy² + dx + ey + f = 0, we must have

f = 0,
a + d + f = 0,
aλ² + bλµ + cµ² + dλ + eµ + f = 0,
a(λ + 1)² + b(λ + 1)µ + cµ² + d(λ + 1) + eµ + f = 0.
Since f = 0, the second, third, and fourth equations simplify to
a + d = 0,                            (1)
aλ² + bλµ + cµ² + dλ + eµ = 0,              (2)
a(λ + 1)² + b(λ + 1)µ + cµ² + d(λ + 1) + eµ = 0.
Subtracting these final two equations, one obtains 2aλ + a + bµ + d = 0, hence by (1),
2aλ + bµ = 0.                (3)

Combining (1), (2), and (3), we find that e = (aλ² − cµ² + aλ)/µ. Thus, the constraints imply that any conic section through A, B, C, and D has the form

a x^{2}-2a{\frac{\lambda}{\mu}}x y+c y^{2}-a x+{\frac{a\lambda^{2}-c\mu^{2}+a\lambda}{\mu}}y=0          (4)

for some real numbers a and c. Such a conic section is an ellipse if and only if its discriminant is negative. Therefore, a point P = (x, y) is in the set if and only if there exist real numbers a and c such that (4) holds and (−2aλ/µ)² − 4ac < 0; this inequality is equivalent to a²λ²/µ² < ac. Clearly, a ≠ 0, and dividing the equation of the conic section by a we get

If we now put c_1 = c/a, we see that P is in the set if and only if there exists a number c_{1} such that

The equality can be rewritten as

c_{1}(y^{2}-\mu y)=-x^{2}+2\,{\frac{\lambda}{\mu}}\,x y+x-{\frac{\lambda^{2}+\lambda}{\mu}}\,y.        (5)

We now distinguish three cases.

Case 1. y² − µy = 0. Then y = 0 or y = µ. If y = 0, the equality yields x = 0 or x = 1, and we find that P = A or P = B.

Similarly, for y = µ we find that P = C or P = D.

Case 2. y² − µy > 0. In this case λ²/µ² < c_1 is equivalent to
(λ^2/µ^2)(y^2 − µy) < c_1(y^2 − µy),
and this together with (5) shows that P is in the set if and only if the following equivalent inequalities hold:

(Note that since µ > 0, it would be impossible to have λy − µx > 0 and λy − µx + µ < 0.)
Now it is easy to check that λy−µx = 0 is the equation of line AD, while λy − µx = −µ is the equation of BC. So the condition −µ < λy − µx < 0 implies that P is between (the extensions of) AD and BC. On the other hand, y² − µy > 0 is equivalent to (y < 0 or y > µ), which says that P is not between AB and CD.

Case 3. y² − µy < 0. The same computation, but with the inequalities reversed, shows that P is not between the lines AD and BC, while P is between (the extensions of) AB and CD. Combining the results of the three cases, we get the answer given above.

Solution 2. Since the set consisting of an ellipse and its interior is convex, it is impossible for a point on an ellipse to lie inside the triangle determined by three other points on that ellipse. We see from this that only the points P described above can be on the same ellipse as A, B, C, and D. For instance, if P is in the region labeled “I” in the figure, then C is inside triangle BDP, so P cannot be on any ellipse on which B, C, and D lie.
To see the converse, choose a coordinate system for the plane such that A = (0, 0), B = (1, 0), C = (λ + 1, µ), and D = (λ, µ), where µ ≠ 0. The linear transformation T taking (x, y) to (x − (λ/µ)y,(1/µ)y) takes A,B, C, and D to (0, 0), (1, 0), (1, 1), and (0, 1), respectively. Since both T~and~T^{−1} map straight lines to straight lines and ellipses to ellipses, we see that it suffices to consider the case A = (0, 0), B = (1, 0), C = (1, 1), and D = (0, 1). In this case the points P = (r, s), other than A, B, C, D, described above lie in two horizontal and two vertical “half strips” given by (r − r²)/(s² − s) > 0. For such a point P, the curve

is an ellipse (in fact, the unique ellipse) through A, B, C, D, and P.