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What are the solutions to the system of conics?

x^2/4 − y^2/2 = 1

x^2 = 4(y+2)

 

 Nov 15, 2019
 #1
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x^2/4 − y^2/2 = 1

x^2 = 4(y+2)        (1)

 

Multiply the first equation through by  4  and we get that

 

x^2 -  2y^2  =  4

x^2  = 2y^2 + 4      (2)

 

Equate (1) and (2)   and we have that

 

4(y + 2)   =  2y^2 + 4

2 (y + 2)  = y^2 + 2

2y + 4  =  y^2 + 2

y^2 - 2y - 2  =  0      complete the square on y

 

 

y^2 - 2y + 1   - 2  - 1   = 0

( y - 1)^2  - 3  =  0

(y - 1)^2  =  3

y - 1  = ±√3

y = 1 ±√3

y = 1 + √3      or  y  = 1 -√3

 

And

x^2  = 4 ( y + 2)

x = ±2√[ y + 2]

 

 

So the solutions are

 

(2√[ 3 - √3] , 1 -√3 )  

 

(-2√[ 3 - √3] , 1 -√3 )

 

(2√[ 3 + √3] , 1 +√3 )

 

(-2√[ 3 + √3] , 1 +√3 )      

 

 

 

cool cool cool

 Nov 15, 2019

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