An arch in the shape of the top half of an ellipse is 10 feet wide and 3 feet high at the center. Find the height of the arch 1 foot from the center.
by just look at the question I advise to first draw it. then use the Ellipse function to determine the height from there
I drew and got the formula of x^2 / 25 + y^2 / 9 =1. I'm not sure where to go from there; would I set up proportions?
To find the height of the arch 1 foot from the center, we can use the equation of an ellipse. The equation of an ellipse with center (h, k), horizontal radius a, and vertical radius b is:
((x-h)^2 / a^2) + ((y-k)^2 / b^2) = 1
In this case, since the arch is symmetrical, the center of the ellipse coincides with the center of the arch. Therefore, the center of the ellipse is (0, 0).
The arch is 10 feet wide, so the horizontal radius of the ellipse, a, is half of that, which is 10/2 = 5 feet.
The height of the arch at the center, which corresponds to k in the equation, is given as 3 feet.
We need to find the height 1 foot from the center, so we can use the equation of the ellipse to solve for y at x = 1.
Substituting the given values, we have:
((1-0)^2 / 5^2) + ((y-0)^2 / b^2) = 1
Simplifying the equation gives:
1/25 + y^2 / b^2 = 1
To solve for y, we need to find the value of b (the vertical radius of the ellipse).
Since the center of the ellipse coincides with the center of the arch, the height of the arch at the center is equal to the vertical radius of the ellipse. Hence, b = 3 feet.
Substituting the value of b, we have:
1/25 + y^2 / 3^2 = 1
Simplifying further:
1/25 + y^2 / 9 = 1
Multiplying by 225 (common denominator) gives:
9 + 25y^2 = 225
25y^2 = 216
Dividing by 25:
y^2 = 216/25
Taking the square root of both sides:
y = ±√(216/25)
y ≈ ±4.15
Since the arch is symmetrical, we take the positive value of y:
y ≈ 4.15 feet
Therefore, the height of the arch 1 foot from the center is approximately 4.15 feet.