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For y > 0, the given curve 2y^2 + x^2 = 64 is moved 5 units to the right.  Find the point of intersection (x,y).

 

 Dec 22, 2019
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When we move this 5 units to the right, the function becomes

 

2y^2  +  (x - 5)^2   =  64

 

Subtracting the given function from this gives us

 

(x - 5)^2  = x^2      simplify

 

x^2 -10x + 25  =   x^2

 

-10x + 25

 

-10x = -25

 

x=  -25/-10   =   2.5   =  the x coordinate of the intersection 

 

Andusing either equation to find y, we have

 

2y^2 + (2.5)^2  =  64

 

2y^2  +   6.25  =  64     subtract    6.25 from both sides

 

2y^2  = 57.75     divide both sides by   2

 

y^2  = 28.875      take the positive root

 

y  = sqrt (28.875)

 

So   ( x , y)   =   (2.5, sqrt (28.875) )

 

See the graph here  :  https://www.desmos.com/calculator/3lkazxxzgq

 

 

cool cool cool

 Dec 22, 2019

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