back to analytical geometry

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.