How do you find the coordinates of the center, foci, lengths of the major and minor axes given (x-1)^2/20+(y+2)^2/4=1?

1 Answer
Nov 19, 2016

Please see the explanation.

Explanation:

Standard form for the equation of an ellipse is:

(x - h)^2/a^2 + (y - k)^2/b^2 = 1

Center: (h,k)

If a > b then:

Foci: (h - sqrt(a^2 - b^2),k) and (h + sqrt(a^2 - b^2),k)
Length of major axis: 2a
Length of minor axis: 2b

If b > a then:

Foci: (h, k - sqrt(b^2 - a^2),k) and (h, k + sqrt(b^2 - a^2))
Length of major axis: 2b
Length of minor axis: 2a

Put the given equation in standard form:

(x - 1)^2/(sqrt(20))^2 + (y - -2)^2/2^2 = 1

Center: (1, -2)

sqrt(20) > 2, therefore, first group applies:

Foci: (1 - sqrt(20 - 4), -2) and (1 + sqrt(20 - 4), -2)

Foci simplified: (-3 , -2) and (5, -2)

Length of major axis: 2sqrt(20)
Length of minor axis: 4