Ellipse Math Example 4

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Example 4

hard
Find the eccentricity of the ellipse (xโˆ’2)236+(y+1)220=1\frac{(x-2)^2}{36} + \frac{(y+1)^2}{20} = 1 and describe what it tells us about the shape.

Solution

  1. 1
    a2=36a^2 = 36, b2=20b^2 = 20. c2=a2โˆ’b2=16c^2 = a^2 - b^2 = 16, so c=4c = 4. Eccentricity e=ca=46=23e = \frac{c}{a} = \frac{4}{6} = \frac{2}{3}.
  2. 2
    Since 0<e<10 < e < 1 and e=23e = \frac{2}{3} is moderately close to 11, the ellipse is noticeably elongated (more oval than circular). A circle has e=0e = 0.

Answer

e=23โ‰ˆ0.667e = \frac{2}{3} \approx 0.667
Eccentricity e=cae = \frac{c}{a} measures how elongated an ellipse is. Values range from 00 (perfect circle) to just below 11 (very elongated). At e=2/3e = 2/3, the ellipse is moderately stretched โ€” the foci are 2/32/3 of the way from center to vertex.

About Ellipse

The set of all points in a plane where the sum of the distances to two fixed points (foci) is constant. Standard form: (xโˆ’h)2a2+(yโˆ’k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.

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