+286 Square ABCD has side length 6. An ellipse $\mathcal{E}$ is circumscribed about the square and there is a point $P$ on the ellipse such that $PC = PD = 8$. What is the area of ellipse $\mathcal{E}$? (You may assume the sides of the square are parallel to the axes of the ellipse.)
+130462 Let A =(-3,3) B = (3,3) C =(3,-3) D = (-3,-3)
Let the center of the ellipse = (0,0) ....let one of the axis lie along x....such that (a,0) = P
The equation of the ellipse is x^2/a^2 + y^2/b^2 =1
Constructing circles at (3,3) and (3,-3) both with radii = 8 we have the equations
(x - 3)^2 + ( y -3)*2 =8^2
(x -3)^2 + (y +3)^2 = 8^2
Let y = 0 and we can find a as
(x-3)^2 + 9 = 8^2
(x -3)^2 = 55
x- 3 = sqrt (55)
x = sqrt(55) + 3 = a ≈ 10.42
To find b^2 we can use this equation
x^2 / 10.42^2 + y^2 / b^2 =1
Since (x,y) = (3,3) is on the ellipse then
3^2 /10.42^2 + 3^2 / b^2 = 1
9 / 10.42^2 + 9/b^2 =1
9 / b^2 = 1 - 9/10.42^2
9/b^2 = .917
b^2 = 9/9.17
b = sqrt (9/.917) ≈ 3.13
The approximate area of the ellipse = pi * a * b = pi * 10.42 * 3.13 = 102.46
