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#?

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Answer 1 · 2016-11-19T21:39:42

Please see the 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