Find properties of an ellipse

Question:

An ellipse has the equation 25x² + 169y² = 4225. Find its major axis, minor axis, foci, eccentricity and vertices.

Solution:

Change the given ellipse equation into the standard ellipse equation.

25x² + 169y² = 4225

both sides of the equation divide by 4225

25x²/4225 + 169y²/4225 = 1

make the coefficient of numerator as one

x²/169 + y²/25 = 1

x²/13² + y²/5² = 1

Compare above equation with the standard ellipse equation x²/a² + y²/b² = 1. (a > b > 0)

we get, a = 13, b = 5

c² = a² - b² = 13² - 5² = 169 - 25 = 144 = 12²

so, c = 12

major axis: 2a = 2 × 13 = 26

minor axis: 2b = 2 × 5 = 10

Foci: F₁ = (-c, 0), so F₁ = (-12, 0)

Foci: F₂ = (c, 0), so F₂ = (12, 0)

eccentricity: e = c/a = 12/13 = 0.92

vertices: (-a, 0), (a, 0), (0, b), (0, -b)

vertices: (-13, 0), (13, 0), (0, 5), (0, -5)

Therefore, the major axis is 26, the minor axis is 10, the foci are (-12, 0) and (12, 0), the eccentricity is 0.92. When c is close to a, b is small, the ellipse is flat. The vertices are (-13, 0), (13, 0), (0, 5), (0, -5). Please watch the video for more details.