Question

Select the correct equations in the image.

Answer 1

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.