SOLUTION: Consider the ellipse x^2/9 + y^2/4 = 1 in the first quadrant.
Lines passing through the point (3/2, 1/4), which is inside this quarter-ellipse,
are free to pivot around it insi
Algebra ->
Surface-area
-> SOLUTION: Consider the ellipse x^2/9 + y^2/4 = 1 in the first quadrant.
Lines passing through the point (3/2, 1/4), which is inside this quarter-ellipse,
are free to pivot around it insi
Log On
Question 1183646: Consider the ellipse x^2/9 + y^2/4 = 1 in the first quadrant.
Lines passing through the point (3/2, 1/4), which is inside this quarter-ellipse,
are free to pivot around it inside this region up to the y- and x-intercepts (0,2) and (3,0), respectively.
What is the maximum or minimum area of the region that can be enclosed by the ellipse x^2/9 + y^2/4 = 1,
any of the pivoting lines around the point (3/2, 1/4), and the x or y axis?
Thank you in advance. Answer by Solver92311(821) (Show Source):
The desired area in each case illustrated below is the area of the quarter ellipse minus the area of the shaded triangle. So, the larger the triangle, the smaller the desired area, and the minimum triangle gives the maximum desired area and vice versa. The area of the quarter ellipse without any subtractions is:
You can use a simple trig substitution to solve this integral. Hint: Take out the constant term and then let and
Then subtract the triangle areas illustrated:
John
My calculator said it, I believe it, that settles it
From
I > Ø
