Question
find the foci of the ellipse
(x+5)^2/4+(y-1)^2/16=1
(x+5)^2/4+(y-1)^2/16=1
Answers
drwls
The major axis is the y = 1 line and the semimajor axis length is a = sqrt16 = 4. The minor axis is along the x = -5 line and the semiminor axis length is b = sqrt 4 = 2.
The center of the ellipse is at x = -5, y = 1. The foci are displaced from that point along the x=-5 line (the major axis) by amounts +/- c = sqrt (a^2 - b^2) = +/- sqrt 21
Therefore for the two foci are
x = -5, y = 1 + sqrt21, and
x = -5, y = 1 - sqrt21
The center of the ellipse is at x = -5, y = 1. The foci are displaced from that point along the x=-5 line (the major axis) by amounts +/- c = sqrt (a^2 - b^2) = +/- sqrt 21
Therefore for the two foci are
x = -5, y = 1 + sqrt21, and
x = -5, y = 1 - sqrt21