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
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.
anyone realize this was asked in 2013?
thankfully this is also relevant 2023! Thanks Kiwi!
Oh, tracks for amusement parks! Now that's a job with some ups and downs! Alright, let's have some fun with math, shall we?
To find the length of the major axis of the track, we need to figure out the distance between the two farthest points on the ellipse along the x-axis. And lucky for us, the equation of the ellipse gives us the clues!
Now, from the equation x^2/2500 + y^2/8100 = 1, we can identify the coefficients of x^2 and y^2. The coefficient of x^2, which we'll call a^2, is 2500. But since we're looking for the length of the major axis, we'll need a, not a^2.
So, taking the square root of 2500 gives us a = 50. Great!
And there you have it—the length of the major axis of the track is 50 yards.
Now, don't get me wrong, this math stuff is a roller coaster ride, but in the end, we always find our way back to the laughter! Enjoy your amusement park design adventure!
To find the length of the major axis of the track, we need to determine the distance between two points on the ellipse's x-axis.
The general equation of an ellipse centered at (h, k) with the semi-major axis length of a and semi-minor axis length of b is:
[(x - h)^2 / a^2] + [(y - k)^2 / b^2] = 1
In this case, the equation of the ellipse is:
x^2/2500 + y^2/8100 = 1
Comparing this with the general equation, we can see that the semi-major axis length (a) is the square root of 2500 and corresponds to the x-coordinate.
So, a^2 = 2500
Taking the square root of both sides, we get:
a = √2500 = 50
Since the major axis length of an ellipse is twice the length of the semi-major axis, the length of the major axis is:
2 * a = 2 * 50 = 100 yards
Therefore, the major axis of the track is 100 yards long.