Asked by lee
Skip designs tracks for amusement parks. For a new design the track will be ellipitical if the ellipse is placed on a large coordinate grid with its center at (0,0) the equation x^2/2500+y^2/8100=1 model the path of the park.The units are given in yards, how long is the major axis of the track?
Please help me
Please help me
Answers
Answered by
Kiwi
This is very late so for people who might need it at some point here's what I did:
The larger denominator is under y so the ellipse is going to be vertical.
Equation: (x-h)^2/b^2 + (y-k)^2/a^2 =1
The center is (0,0) right so both h and k will be 0
a is the length of the major axis
b is the length of the minor axis
x^2/50^2 + y^2/90^2 = 1
x^2/50^2 + y^2/90^2 = 1
a=90
90 x 2 is 180
The major axis' length is going to be 180 yards.
The larger denominator is under y so the ellipse is going to be vertical.
Equation: (x-h)^2/b^2 + (y-k)^2/a^2 =1
The center is (0,0) right so both h and k will be 0
a is the length of the major axis
b is the length of the minor axis
x^2/50^2 + y^2/90^2 = 1
x^2/50^2 + y^2/90^2 = 1
a=90
90 x 2 is 180
The major axis' length is going to be 180 yards.
Answered by
a Human
thankfully this is also relevant 2023! Thanks Kiwi!
Answered by
A Human
anyone realize this was asked in 2013?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.