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# help

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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.

Dec 6, 2019

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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   Dec 6, 2019