an ellipse has an area of 4.17 in^2, and a minor axis that is 2 inch long. solve fot the major axis.
how do i solve this?
for any ellipse of the form
x^2/a^2 + y^2/b^2 = 1
the area of the ellipse is (ab)pi
so if the minor axis is 2 inches, then b=1
abpi = 4.17
api = 4.17
a = 4.17/pi = 1.32735
so the major axis or 2a is 2.6547 units long.
check with this:
http://www.csgnetwork.com/areaellipse.html
To solve for the major axis of an ellipse given the area and the length of the minor axis, you can use the formula for the area of an ellipse:
Area = π * (major axis/2) * (minor axis/2)
In this case, you are given the area (4.17 in^2) and the length of the minor axis (2 inches).
Let's substitute the given values into the formula and solve for the major axis:
4.17 in^2 = π * (major axis/2) * (2 in/2)
To simplify, we can cancel out the common factors:
4.17 = π * (major axis/2)
Now, to solve for the major axis, we need to isolate it by dividing both sides of the equation by π/2:
4.17 / (π/2) = major axis
To get the value of the major axis, divide 4.17 by π/2:
major axis ≈ 1.325 inches
Therefore, the major axis of the ellipse is approximately 1.325 inches.