Answer:
equations 1, 2, 6 . . . (top two, bottom right)
Step-by-step explanation:
For the standard-form equation of an ellipse:
Ax^2 +Bx +Cy^2 +Dy +E = 0
we can define ...
p = min(A, C)
q = max(A, C)
Then the eccentricity can be shown to be ...
e = √(1 -p/q)
p/q = 1 -e^2
For eccentricity < 0.5, we want ...
p/q > 3/4
__
Checking the values of p/q for the given equations left-to-right, top-to-bottom, we have p/q = ...
- 49/64 ≈ 0.765 . . . e < 0.5
- 81/100 ≈ 0.810 . . . e < 0.5
- 6/54 ≈ 0.111 . . . e > 0.5
- 36/49 ≈ 0.734 . . . e > 0.5
- 4/25 ≈ 0.160 . . . e > 0.5
- 64/81 ≈ 0.790 . . . e < 0.5
Equations 1, 2, 6 are of ellipses with eccentricity < 0.5.
