Question

How do I find the length of the major and minor axis of (x-5)^2/4 + (y+3)^2/16=1?

Answer 1

I got that the center is (5,-3)
Vertices = (5,1) (5,-7)
Foci = (5,-3+2√3) (5,-3-2√3) but I cannot find the length of the major and minor axis or the eccentricity...

Answer 2

Do you know what they are?

Answer 3

no, I don't :(

Answer 4

Well if I'm not mistaken, this is an ellipse right?

Answer 5

So that means for the axis:
Major axis: The longest diameter of an ellipse.
Minor axis: The shortest diameter of an ellipse.

Answer 6

yes, you are correct! but how do I find what the major axis and minor axis is?

Answer 7

a^2 is ALWAYS the bigger denominator.
2a = major axis
2b = minor axis
a^2 = 16
a =4
2a = major axis
2(4) = 8
major axis is 8
minor axis = 2b
b^2 = 4
b = 2
minor axis = 4

Answer 8

ahhh, I finally see what you did, thats a lot easier than I thought! how do I find the eccentricity? I know you had said it was like c/a, but what number represents c?

Answer 9

Like in a triangle
but the formula is.
a^2 = b^2 + c^2
so
16 = 4 + c^2
12 = c^2
2√3 = c
eccentricity = c/a

2√3/ 4
= √3/2

Answer 10

thank you SOOO much! you are a life saver! How did you know that c^2 was 12 though?

Answer 11

a^2 = b^2 + c^2
i just plugged in your values.

Answer 12

a^2 = 16
b^2 = 4
and i solved for c :)

Answer 13

oh okay! :)