Question

find the curve of intersection of the elliptic cyclinder x^2 +4z^2 = 16 and the plane x+3y-2z = 3

Answer 1

One thing you could possibly do is multiply the plane by (16/3) giving:
(16/3)x+16y-(32/3)z=16
Now they are both=16
x^2+4z^2=(16/3)x+16y-(32/3)z
Then solve it however you want it.
You could also solve the first equation for z or x but the problem is you have a sqrd so you end up with a +/- ambiguity.

Answer 2

Parametrize the curve of intersection. The elliptic cylinder can be parametrized by
x=4cos(t)
z=2sin(t)
(y=y)
Then rewrite the equation of the plane as y=1+2/3z-1/3x, so
y=1+4/3sin(t)-4/3cos(t)

The curve is then parametrized by
x=4cos(t)
y=1+4/3sin(t)-4/3cos(t)
z=2sin(t)
t ranges from 0 to 2pi