Ellipse with a=13, b=5: find c and e

Given an ellipse with foci on the x-axis, semi-major axis is 13 and semi-minor axis is 5, find its semi-focal distance and eccentricity.

Step-by-step

1. Ellipse: a = 13, b = 5
   The semi-major axis and semi-minor axis of an ellipse determine its size and flatness.
2. c² = a² − b² = 13² − 5² = 144
   In an ellipse, a is the largest, so we compute a² minus b².
3. c = √144 = 12
   Take the positive square root; the semi-focal distance is the distance from a focus to the center.
4. e = c / a = 12 / 13 = 0.923077
   Eccentricity measures flatness; an ellipse has eccentricity between 0 and 1
5. Back-substitution check: a² = b² + c² ⇒ 169 = 25 + 144
   Use the relationship among a, b, and c to back-check that the semi-focal distance is correct.

Why it works

In an ellipse, the semi-major axis is the largest, so the relationship among the three quantities is that the square of the semi-major axis equals the sum of the squares of the semi-minor axis and the semi-focal distance. Rearranging gives the semi-focal distance. The semi-focal distance is the distance from a focus to the center, and the eccentricity is its ratio to the semi-major axis; one quantity describes size while the other describes flatness. After calculating, substitute back to verify whether the three quantities satisfy the relationship.

Common mistakes

• Wrote the relationship as the square of the semi-focal distance equals the square of the semi-major axis minus the square of the semi-minor axis, then rearranged incorrectly.
• Calculated eccentricity as semi-major axis divided by semi-focal distance, taking the reciprocal.
• Forgot to take the positive root when extracting the square root, writing a negative semi-focal distance.

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