Question

Can someone please help me find the lengths of the major and minor axes for:

(x-5)^2/4 + (y+3)^2/16 =1

Answer 1

https://www.google.com/search?q=(x-5)%5E2%2F4+%2B+(y%2B3)%5E2%2F16+%3D&oq=(x-5)%5E2%2F4+%2B+(y%2B3)%5E2%2F16+%3D&aqs=chrome..69i57.290j0j1&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8

Answer 2

One form of an equation for ellipse is:
(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1
where a and b are half of the major/minor axes. Take the bigger one to be the major axis and the smaller one to be the minor axis.
compare (x-5)^2/4 + (y+3)^2/16 =1 to
(x-h)^2 / a^2 + (y-k)^2 / b^2 = 1
a^2 = 4; so a = 2
b^2 = 16; b = 4
a and b are half the length of the axis.
so length of major axis = 2 * b = 2 * 4 = 8
length of minor axis = 2 * a = 2 * 2 = 4