+309 What are the solutions to the system of conics?
x^2/4 − y^2/2 = 1
x^2 = 4(y+2)
+128475 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 )
