Posted by Tom on Friday, November 9, 2012 at 1:50pm.
from the 1st: y^2 = 4x^2 + 7
into the 2nd
9(4x^2 + 7) + 16x^2 = 323
36x^2 + 63 + 16y^2 = 323
52y^2 = 260
y^2 = 5
y = ± √5
sub y^2 = 5 into the 1st:
5 - 4x^2 = 7
-4x^2 = -2
x^2 = 1/2
x = ± 1/√2
4 solutions:
x=√5, y= 1/√2
x = -√5, y = 1/√2
x = √5, y = -1/√2
x = -√5 , y = -1/√2
your graph would show an ellipse intersecting with a hyperbola, with 4 symmetric points of intersection
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