The points of intersection of the graphs of xy = 20 and x^2 + y^2 = 41 are joined to form a convex quadrilateral. Find the area of the quadrilateral.
+128408 xy = 20 ⇒ y =20/x
Sub this into the second circular equation and we have that
x^2 + (20/x)^2 = 41
x^2 + 400/x^2 = 41 multiply through by x^2
x^4 + 400 = 41x^2 rearrange as
x^4 -41x^2 + 400 = 0 factor as
( x^2 - 25) ( x^2 - 16) = 0
So we can find the xcoordinates of the intersections as follows
x^2 - 25 = 0 and x^2 - 16 = 0
x^2 = 25 x^2 =16
x = 5 , x = -5 x = 4, x = -4
Using xy = 20
When x = 5, y =4
When x = -5, y = -4
When x = 4, y =5
When x = -4, y =-5
So....we can nmae the points as
A = (4,5) B = ( 5,4) C = (-4, -5) D = (-5, -4)
This forms a rectangle
One side is AB = sqrt [ (5 - 4)^2 + (5 -4)^2 ]= sqrt [ 1 + 1 ] = sqrt (2)
Another side is BC =sqrt [ (-4 -5)^2 + ( -5 - 4)^2 ] = sqrt [ 9^2 + 9^2] = sqrt [ 2 * 81] = 9sqrt (2)
So the area = AB * BC = sqrt (2) * 9 sqrt (2) = 9 * 2 = 18 units^2
Here's the graph : https://www.desmos.com/calculator/9cnh4yz2ho
