Square ABCD has side length 60. An ellipse E is circumscribed about the square and there is a point P on the ellipse such that PC = PD =50. What is the area of E?
I got to the part with the point (30, 30). Now what next?
If you draw the figure, you will see that the ellipse has semi-major axis of 30+40=70
So, the equation of the ellipse is
x^2/70^2 + y^2/b^2 = 1
Since (30,30) is on the ellipse,
30^2/70^2 + 30^2/b^2 = 1
b = 21√(5/2)
The area of the ellipse is thus
A = πab = 735π√10
To find the area of the ellipse E, we first need to determine the semi-major and semi-minor axes.
Given that square ABCD has a side length of 60, the diagonal of the square is √(60^2 + 60^2) = √(2 * 60^2) = √(2) * 60.
Since the diagonals of a square are also the diameters of the circumscribed ellipse, the length of the semi-major axis (a) of the ellipse is equal to half of the length of the diagonal of the square, which is (√(2) * 60)/2 = √2 * 30.
Next, we need to find the length of the semi-minor axis (b) of the ellipse. Since the square is inscribed in the ellipse, the length of the semi-minor axis is equal to half of the side length of the square, which is 60/2 = 30.
Now that we have the values of a and b, we can use the formula to calculate the area of the ellipse: Area = π * a * b.
Substituting the values, we have Area = π * √2 * 30 * 30 = 900π√2.
Therefore, the area of the ellipse E is approximately 900π√2.